Topics

Instantaneous Velocity:

Instantaneous Velocity is defined as the value approached by the average velocity when the time interval for measurement becomes closer and closer to zero, i.e. \(\Delta t \rightarrow 0\). Mathematically

\[v(t) = \lim_{\Delta t \rightarrow 0}\mspace{2mu}\frac{\Delta x}{\Delta t}\]

Thus Instantaneous velocity function is the derivative of the displacement with respect to time.

\[v(t) = \frac{dx(t)}{dt}\]

Instantaneous Speed

It is measure of how fast a particle or a body is moving at a particular instant. It is the magnitude of instantaneous velocity. Thus particle moving with instantaneous velocity of \(+ 5\text{ }m/s\) and another moving with \(- 5\text{ }m/s\) will have same instantaneous speed of \(5\text{ }m/s\).
The speedometer in a car measure the instantaneous speed not the instantaneous velocity, because it cannot determine the direction.

Average Acceleration

• For any change in velocity either in its magnitude or direction or both, acceleration must be present.

Without acceleration neither direction nor magnitude of velocity can be changed.
When a particle's velocity changes, the particle is said to undergo acceleration (or to accelerate).
The Average Acceleration ( \({\overrightarrow{\mathbf{a}}}_{\text{avg}\text{~}}\) ) over a time interval \(\Delta t\) is

\[{\overrightarrow{\mathbf{a}}}_{\text{avg}\text{~}} = \frac{{\overrightarrow{v}}_{2} - {\overrightarrow{v}}_{1}}{t_{2} - t_{1}} = \frac{\Delta\overrightarrow{v}}{\Delta t}\]

where the particle has velocity \(v_{1}\) at time \(t_{1}\) and then velocity \(v_{2}\) at time \(t_{2}\).

Instantaneous Acceleration

The Instantaneous Acceleration (or simply acceleration) is the derivative of the velocity with respect to time.

\[\overrightarrow{\mathbf{a}} = \frac{d\overrightarrow{v}}{dt}\]

In words, the acceleration of a particle at any instant is the rate at which its velocity is changing at that instant.

\[\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} = \frac{d}{dt}\left( \frac{\text{ }d\overrightarrow{r}}{dt} \right) = \frac{d^{2}\overrightarrow{r}}{{dt}^{2}}\]

In words, the acceleration of a particle at any instant is the second derivative of its position vector with respect to time.
Acceleration has both magnitude and direction (it is yet another vector quantity). For motion on a straight line its algebraic sign represents its direction on an axis just as for displacement and velocity; that is, acceleration with a positive value is in the positive direction of an axis, and acceleration with a negative value is in the negative direction.

Note :

(1) Here we can observe at \(t = 0.4\text{ }s\), particle has zero velocity but acceleration of \(10\text{ }m/s^{2}\). Thus particle having zero velocity need not have zero acceleration.
(2) For \(t < 0.4\text{ }s\), velocity is negative and for \(t > 0.4\text{ }s\), velocity is in positive direction i.e. its velocity
changes its direction at \(t = 0.4sec\), when becoming zero.

Practice Exercise

Q. 1 If a particle traverses on a semicircular path of radius R from A to B as shown in time T , find average speed and average velocity.

Q. 2 A man runs for first 120 m at \(6\text{ }m/s\) and then next 120 m at \(3\text{ }m/s\) in the same direction. Find
(a) Total time of run
(b) Average velocity
Q. 3 Position of particle moving along x -axis is given by
\(x = \left( 3t^{2} - 2t^{3} \right)m\ (t\) is in sec \()\)
Find
(a) its average velocity form \(t = 0\text{ }s\) to \(t = 2\text{ }s\)
(b) \(v(t)\) and a (t)
(c) The time at which its acceleration is zero and find velocity at the instant.
Q. 4 Position of a particle moving along straight line is given by \(x = \left( - t^{2} + 6t + 5 \right)m\) ( t is in sec) Find
(a) The time at which velocity of particle is zero.
(b) Average velocity from \(t = 0\) to \(t = 4sec\)
(c) Average speed from \(t = 0\) to \(t = 4sec\)

Answers

1. Average speed \(= \frac{\pi R}{T};\ \) Average velocity \(= \frac{2R}{T}\) (from A to B)

2. (a) \(60\text{ }s\ \) (b) \(4\text{ }m/s\ 3\). (a) \(- 2\text{ }m/s\ \) (b) \(v(t) = 6t - 6t^{2}\text{ }m/s\) (c) \(\frac{1}{2}\text{ }s;\frac{3}{2}\text{ }m/s\)

3. (a) \(3sec\ \) (b) \(- 2\text{ }m/s\ \) (c) \(2.5\text{ }m/s\).

Motion with variable acceleration

Relations :

(i) \(\ \frac{dv}{dt} = a\ \Rightarrow \ \int_{v_{1}}^{v_{2}}\mspace{2mu} dv = \int_{t_{1}}^{t_{2}}\mspace{2mu} adt\)
(iii) \(\ \frac{dx}{dt} = v\ \Rightarrow \ \int_{x_{1}}^{x_{2}}\mspace{2mu} dx = \int_{t_{1}}^{t_{2}}\mspace{2mu} vdt\)
(iii) \(a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}\ \) (By chain rule)

\[\begin{matrix} \therefore & a = v\frac{dv}{dx} \\ \therefore & \int_{v_{1}}^{v_{2}}\mspace{2mu}\mspace{2mu} Vdv = \int_{x_{1}}^{x_{2}}\mspace{2mu}\mspace{2mu} adx \end{matrix}\]

Graphical Representation of Motion in one Dimension

S-t curve :

If we put \(\mathbf{s}\) on y -axis and \(\mathbf{t}\) on x -axis then for every value of \(\mathbf{t}\) we have a specific value of \(\mathbf{s}\).

The Average velocity from time \(t_{1}\) to \(t_{2}\) will be

\[V_{\text{avg}\text{~}} = \frac{s_{2} - s_{1}}{t_{2} - t_{1}} = \text{~}\text{slope of line joining the points}\text{~}p_{1}\text{~}\text{and}\text{~}p_{2}\]

For a particle moving along a straight line when we plot a graph of \(\mathbf{s}\) versus \(\mathbf{t},\mathbf{V}_{\text{avg}\text{~}}\) is the slope of the straight line that connects two particular points on the \(\mathbf{s}(\mathbf{t})\) curve : one is the point that corresponds to \(s_{2}\) and \(t_{2}\), and the other is the point that corresponds to \(s_{1}\) and \(t_{1}\). Like displacement, \(\mathbf{v}_{\text{avg}\text{~}}\) has both magnitude and direction (it is another vector quantity). Its magnitude is the magnitude of the line's slope. A positive \(\mathbf{v}_{\text{avg}\text{~}}\) (and slope) tells us that the line slants upward to the right ; a negative \(v_{\text{avg}\text{~}}\) (and slope), that the line slants downward to the right.

Instantaneous velocity

According to definition

\[v = \lim_{\Delta t \rightarrow 0}\mspace{2mu}\frac{\Delta s}{\Delta t}\]

In curve if \(\Delta t \rightarrow 0\) the point \(p_{2}\) comes very close to point \(p_{1}\).

Note: The instantaneous velocity can be found by determining the slope of the tangent to the displacement time graph at that instant. Velocity at point \(p_{1}\) or time \(t_{1}\) is V

\[V = tan\theta\]

Cases :

(A) Uniform velocity :

If velocity is uniform slope of curve must remain unchanged. Curve with uniform slope is straight line
e.g. (i) \(S = Vt\), If Velocity is \(1{\text{ }ms}^{- 1} \Rightarrow \text{ }S = t\)

\(tan\theta = 1\)
e.g. (ii) If velocity is \(- 1\text{ }m/s \Rightarrow S = - t\)

\[tan\theta = - 1\]

For negative velocity

(B) Uniform acceleration

We have a particle moving with uniform acceleration a and initial velocityu. Its displacement s at any time t can be represented as

\[s = ut + \frac{1}{2}{at}^{2}\]

Curve is parabola
Velocity at \(t_{1}\) is \(tan\theta\)

Velocity Vs time curve

By using dependence of \(v\) on \(t\) we can plot a Vst graph.
Slope of v Vs t curve at any point represents acceleration at that instant.

\[tan\theta = \text{~}\text{acceleration at time}\text{~}t_{1}\]

Area under v Vs t graph:

As we know \(\ dx = Vdt\)

\[\begin{matrix} \Rightarrow \ \int_{x_{1}}^{x_{2}}\mspace{2mu}\mspace{2mu} dx & \ = \int_{t_{1}}^{t_{2}}\mspace{2mu}\mspace{2mu} vdt \\ \Rightarrow \ \Delta x & \ = \int_{t_{1}}^{t_{2}}\mspace{2mu}\mspace{2mu} vdt \\ & \ = \text{~}\text{Area under}\text{~}vVst\text{~}\text{graph.}\text{~} \end{matrix}\]

Thus area under curve will represent displacement in that time period.
Note : (1) Area above \(t\)-axis +ve displacement.
(2) Area below t-axis is-ve displacement.

Thus,

1. Total displacement will be sum of areas with appropriate signs.

2. Total distance travelled will be sum of areas without sign.

Cases :
(1) For uniform velocity:
acceleration \(= 0\)
slope \(= 0\)

(2) For uniform st. line curve
\(tan\theta =\) acceleration
For increasing velocity

\(tan\theta =\) acceleration
for decreasing velocity
(slope is-ve) i.e. \(\theta > 90^{\circ}\)

Note \(\theta\) is always with + ive \(x\)-axis

Table for :
Variation of Displacement (s), velocity (v) and acceleration (a) with respect to time for different type of motion,

1. At rest		Velocity	Acceleration
2. Motion with constant velocity
3. Motion with constant acceleration
4. Motion with constant deceleration

Practice Exercise

Q. 1 A particle starts moving with speed \(3\text{ }m/s\) and accelerates for 5 sec . with acceleration \(2\text{ }m/s^{2}\). Find the displacement of the particle.
Q. 2 A particle has an initial velocity of \(9\text{ }m/s\) due east and has a constant acceleration of \(2\text{ }m/s^{2}\) due west. Find the distance covered by the particle in the \(5^{\text{th}\text{~}}\) second of its motion.
Q. 3 The acceleration of a particle traveling along a straight line is shown in the figure. What is the maximum speed of the particle?

Q. 4 A runner is at the position \(x = 0\text{ }m\) when time \(t = 0\text{ }s\). One hundred meters away is the finish line. Every ten seconds, this runner runs half the remaining distance to the finish line. During each ten-second segment, the runner has a constant velocity. For the first thirty seconds of the motion, construct
(a) the position-time graph.
(b) the velocity-time graph.
Q. 5 Velocity of particle starting from rest varies with position according to equation \(v = \sqrt{\alpha x}\). What is distance travelled by particle in \(t\) second from start?
Q. 6 A body starts from origin and moves along \(x\)-axis such that at any instant velocity is \(v = 4t^{3} - 2t\). Find the acceleration of the particle when it is 2 m from the origin.

Answers
Q. 1	2 sec	Q. 2	0.5 m	Q. 3	\[30\text{ }m/s\]
Q. 4
Q. 5	\[\frac{1}{4}\alpha t^{2}\]	Q. 6	\[22\text{ }m/s^{2}\]

Vertical motion under gravity (Free fall)

Motion that occurs solely under the influence of gravity is called free fall. Thus a body projected upward or downward or released from rest are all under free fall.
In the absence of air resistance all falling bodies have the same acceleration due to gravity, regardless of their sizes or shapes.
The value of the acceleration due to gravity depends on both latitude and altitude. It is approximately \(9.8\text{ }m/s^{2}\) near the surface of the earth. For simplicity a value of \(10\text{ }m/s^{2}\) is used. To do calculations regarding motion under gravity, we follow a proper sign convention. We are taking upward direction as positive and downward as negative, Thus acceleration is taken \(a = - g = 10\text{ }m/s^{2}\) no matter whether body is moving upwards or downwards, since \(\mathbf{g}\) always acts downward.
Thus the equation of kinematics may be modified as

\[\begin{array}{r} v = u - gt\#(i) \end{array}\]

\[\begin{array}{r} \Delta y = y - y_{0} = ut - \frac{1}{2}{gt}^{2}\#(ii) \end{array}\]

\[\begin{array}{r} v^{2} = u^{2} - 2\text{ }g\left( y - y_{0} \right)\#(iii) \end{array}\]

These \(y_{0} =\) position of particle at time \(t = 0\)
\(y =\) position of particle at time \(t\).
\(u =\) velocity of particle at time \(t = 0\)
\(v =\) velocity of particle at time t .

Some results

1. Maximum Height :- \(\mathbf{H} = \frac{\mathbf{u}^{2}}{\mathbf{2g}}\)

Derivation : At maximum height \(v = 0\)
\(\therefore\ \) from equation (iii), \(v^{2} = u^{2} - 2gH = 0 \Rightarrow H = \frac{u^{2}}{2\text{ }g}\)
2. Time to reach maximum height : \(- \mathbf{t} = \frac{\mathbf{u}}{\mathbf{g}}\)

Derivation : At maximum height \(v = o = u - gt\) [equation (i)]
\[\therefore\ t = \frac{u}{g} \]3. Total time of flight = time to go up + time to move down (to reach the same horizontal level again)

\[\begin{matrix} T & \ = 2t \\ \mathbf{T} & \ = \frac{\mathbf{2u}}{\mathbf{g}} \end{matrix}\]

4. Time of ascent = Time of descent for motion between two points at same horizontal level for example between A & B and between C & D shown in the figure.

5. If an object is dropped ( means initial velocity is zero) from Height \(h\). Its speed on reaching ground is \(V = \sqrt{2gh}\) and time taken to reach ground is \(t = \sqrt{\frac{2\text{ }h}{\text{ }g}}\)

Derivation : From equation (iii) \(0 - 2g( - h) = v^{2}\ \lbrack\therefore\Delta y = - h\rbrack\)
Also from equation (ii) \(\Delta y = - h = 0 - \frac{1}{2}{gt}^{2}\)
\[\therefore\ t = \sqrt{\frac{2\text{ }h}{\text{ }g}} \]6. A particle has the same speed at a point on the path. While going vertically up and down.

Motion In Two Dimensions

Whatever we have studied in kinematics of one dimensional motion, we apply the same for motion in two and three dimensional motion, for \(x,y\) & Z components separately.
Suppose a particle has position coordinates \((x,y)\) at any instant, then its position vector is
given by, \(\ \overrightarrow{r} = x\widehat{i} + y\widehat{j}\)
If particle moves from point A to B , through any path, then its displacement is

\[\Delta\overrightarrow{r} = {\overrightarrow{r}}_{2} - {\overrightarrow{r}}_{1}\]

\[= \Delta x\widehat{i} + \Delta y\widehat{j}\]

Now at any instant, its velocity is given by

\[\overrightarrow{V} = \frac{d\overrightarrow{r}}{dt} = \left( \frac{dx}{dt} \right)\widehat{i} + \left( \frac{dy}{dt} \right)\widehat{j}\]

i.e \(\ V_{x} = \frac{dx}{dt}\ \) i.e. \(x\) - component of velocity.

and \(\ V_{y} = \frac{dy}{dt}\ \) i.e. \(y\) - component of velocity
Similarly \(\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} = a_{x}\widehat{i} + a_{y}\widehat{j}\)
Where \(a_{x} = \frac{{dv}_{x}}{dt}\& a_{y} = \frac{{dv}_{y}}{dt}\)

Projectile Motion

It consists of two independent motions, a horizontal motion at constant velocity and a vertical motion under acceleration due to gravity.
In order to deal with problems in projectile motion, one has to choose a coordinate system. Let's take horizontal as x -axis and vertical upward direction as y -axis, then \(\mathbf{a}_{\mathbf{x}} = \mathbf{0}\) and \(\mathbf{a}_{\mathbf{y}} = - \mathbf{g}\); since there is only one force " mg " downward (negative air resistance)

Equation along \(x\)-axis
\(v_{x} = u_{x}\) ( constant)
\[\Delta x = u_{x}t\]

Equations along \(y\)-axis
\[{v_{y} = u_{y} - gt }{\Delta y = u_{y}t - \frac{1}{2}{gt}^{2} }{v_{y}^{2} = u_{y}^{2} - 2\text{ }g(\Delta y)}\]

If an object is dropped from rest or projected up or down, it follows straight line path. If its initial velocity is not along the line of force it follows parabolic path which is proved mathematically in this topic later on.

Projectile Thrown from the Ground Level

A particle is projected from ground level at an angle \(\theta\) from horizontal with speed \(u\).

22
\(\therefore u_{x} = ucos\theta\ \) and \(u_{y} = usin\theta\)
At any instant, \(v_{x} = u_{x} = ucos\theta\& v_{y} = u_{y} - gt = usin\theta - gt\)

\[\Delta x = (ucos\theta)t\ \&\ \Delta y = (usin\theta)t - \frac{1}{2}gt^{2}\]

Time of flight (T): Let it strikes the ground again at time T.
i.e. for \(t = T,\Delta y = 0u_{y} = 0 = u_{y}T - \frac{1}{2}{gT}^{2}\)
\[\therefore\ T = \frac{2u_{y}}{g} = \frac{2usin\theta}{g} \]Horizontal Range ( \(\mathbf{R}\) ) It is horizontal displacement till time \(t = T\)
i.e. \(\ R = u_{x}T\)

\[R = \frac{2u_{x}u_{y}}{\text{ }g}\]

i.e. \(R = \frac{2(ucos\theta)(usin\theta)}{g} = \frac{u^{2}sin(2\theta)}{g}\)

Maximum height (H)

\[H = \Delta y\ \text{~}\text{when}\text{~}v_{y} = 0\]

\[{\therefore\ v_{y}\ ^{2} = u_{y}\ ^{2} - 2\text{ }g(H) = 0 }{\therefore\ H = \frac{u_{y}^{2}}{2\text{ }g} = \frac{u^{2}\sin^{2}\theta}{2\text{ }g} }\]From above formulae, we can observe
(i) \(T \propto u_{y}\) i.e. depends only on vertical component of initial velocity
(ii) \(H \propto u_{y}\ ^{2}\) i.e. depends only on vertical component of initial velocity
(iii) \(R \propto u_{x}u_{y}\) i.e. depends both on horizontal and vertical components of initial velocity.

Velocity at any general point

\[v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(ucos\theta)^{2} + (usin\theta - gt)^{2}} = \sqrt{u^{2} + g^{2}t^{2} - 2(usin\theta)gt}\]

If angle which direction of motion makes at an instant is \(\phi\), then

\[tan\phi = \frac{v_{y}}{v_{x}} = \frac{usin\theta - gt}{ucos\theta}\]

\(tan\phi\) is positive during its upward notion i.e. before reaching highest point and after that \(tan\phi\) is negative.

Equation of Trajectory :

\[y = usin\theta t - (1/2){gt}^{2}\]

and

\[x = (ucos\theta)t\ \Rightarrow \ t = \frac{x}{ucos\theta}\]

From these equations, (eliminating t)

\[\mathbf{y} = \mathbf{x}tan\theta - \frac{g}{2u^{2}\cos^{2}\theta}\mathbf{x}^{2}\]

The above relation between x and y is equation of parabola, which proves that the trajectory i.e. path of projectile is parabolic.

Other points of remember

• \(R = \frac{u^{2}sin(2\theta)}{g}\)

Range is maximum, when \(\theta = 45^{\circ}\)
and \(\ R_{\max} = \frac{u^{2}}{\text{ }g}\)

• For two objects projected with same speed Range is same for two angles of projection \(\theta\) & (90- \(\theta\) )

Proof: Let \(R_{1} = R_{2}\) for \(\theta\) and \(\alpha\)
i.e. \(\frac{u^{2}sin(2\theta)}{g} = \frac{u^{2}sin(2\alpha)}{g}\)
i.e. \(sin(2\theta) = sin(2\alpha)\)
i.e. \(2\theta = 180^{\circ} - 2\alpha\)
\[\Rightarrow \alpha = 90^{\circ} - \theta\]

• \(H = \frac{u^{2}\sin^{2}\theta}{\text{ }g}\) is maximum i.e. \(\left( \frac{u^{2}}{2\text{ }g} \right)\) if projected at \(\theta = 90^{\circ}\)

Horizontal Projection :

In Horizontal Direction

(i) Initial velocity \(u_{x} = u\)
(ii) Acceleration \(= 0\)
(iii) Horizontal velocity of particle remains same after time \(t\) horizontal velocity \(= v_{x} = u\)
(iv) Range \(x = u\) t

In Vertical Direction

(i) Initial velocity \(u_{y} = 0\)
(ii) Acceleration = ' g ' downward
(iii) Velocity of particle after time t \(v_{y} = 0 + ( - g)t = - gt = gt\) (downward)
(iv) Displacement \(y = (1/2){gt}^{2}\) (downward)

Velocity at a general point \(\mathbf{P}(\mathbf{x},\mathbf{y})\) :
\[v = \sqrt{v_{x}^{2} + v_{y}^{2}}\ tan\phi = \frac{v_{y}}{v_{x}} \phi\] is angle made by v with horizontal in clockwise direction

Time of flight:

\[\begin{matrix} & \ - h = u_{y}t - (1/2){gt}^{2} = 0 - \frac{1}{2}{gt}^{2} \\ & t = \pm \sqrt{\frac{2\text{ }h}{\text{ }g}} \\ & t = + \sqrt{\frac{2\text{ }h}{\text{ }g}} \end{matrix}\]

(negative time is not possible)

Range:

\[R\ = u_{x}t = u\sqrt{\frac{2\text{ }h}{\text{ }g}}\]

Note :

If a projectile is projected with initial velocity \(u\) and another particle is dropped from same height at the same time, both the projectile would strike the ground at the same instant velocity. Both will have same vertical components of velocity but their net velocities would be different.

Projected from some height at some angle

Case I:

When projected at some angle \(\theta\) with the horizontal towards upward direction.
Let it takes time \(\mathbf{t}_{\mathbf{1}}\) (time of flight) to strike the ground

\[\begin{matrix} & \Delta y = - h\ \text{~}\text{When}\text{~}t = \mathbf{t}_{1} \\ \therefore & - h = (usin\theta)t_{1} - \frac{1}{2}{gt}_{1}^{2} \\ \Rightarrow & t_{1}^{2} - \left( \frac{2usin\theta}{\text{ }g} \right)t_{1} - \frac{2\text{ }h}{\text{ }g} = 0 \\ \therefore & t_{1} = \frac{T + \sqrt{T^{2} + 8\text{ }h/g}}{2} \\ & \left( \text{~}\text{where}\text{~}T = \frac{2usin\theta}{\text{ }g} \right) \end{matrix}\]

Also \(\ \mathbf{R} = \Delta x = u_{x}t_{1}\) (putting values of \(t_{1}\& u_{x}\) we can find \(\mathbf{R}\) whenever required)
When it reaches ground \(v_{x} = ucos\theta\)

\[\begin{matrix} & \& v_{y}^{2} = u_{y}^{2} - 2g(\Delta y) \Rightarrow v_{y} = \sqrt{(usin\theta)^{2} - 2g( - h)} \\ \therefore & v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{u^{2} + 2gh} \end{matrix}\]

Case II :

When projected at angle \(\theta\) with horizontal towards downward direction
Here \(\ u_{y} = - usin\theta\)
Thus, if it takes time
\(\mathbf{t}_{2}\) to strike the ground then

\[h = - (usin\theta)t_{2} - \frac{1}{2}{gt}_{2}\ ^{2}\]

\[{\Rightarrow \ t_{2} + \left( \frac{2usin\theta}{g} \right)t_{2} - h = 0 }{\therefore\ \mathbf{t}_{2} = \frac{\sqrt{T^{2} + 8\text{ }h/g} - T}{2} }\]

Also \(\ \mathbf{R} = u_{x}t_{2}\)
Here on reaching ground, \(v_{x} = ucos\theta\)
and \(\ v_{y} = \sqrt{(usin\theta)^{2} - 2\text{ }g( - h)}\)
\[\therefore\ v = \sqrt{v_{y}^{2} + v_{x}^{2}} = \sqrt{u^{2} + 2gh} \]Thus we can observe if some particles are projected from same height with same speed, they reach the ground with same speed whatever may be the angles of their projection.

Projectile thrown On An Inclined plane

To deal with problems of projectile thrown along an incline we choose the x -axis along the plane and \(y\)-axis perpendicular to the plane.
Let a particle is projected at an angle \(\alpha\) with the horizontal on an incline plane which has angle of inclination \(\beta\) with the horizontal

(i)

From fig (i) ; \(u_{x} = ucos(\alpha - \beta);u_{y} = usin(\alpha - \beta)\)
\(\lbrack\therefore\) angle of projection with the incline is \((\alpha - \beta)\rbrack\)
From fig(ii) ; \(a_{x} = - gsin\beta;a_{y} = - gcos\beta\)
(a) Time of flight (T) on the incline

At \(t = T\), it strikes the incline
i.e. \(\ \Delta y = 0\)

\[u_{y}\text{ }T + \frac{1}{2}a_{y}{\text{ }T}^{2} = 0\]

\[{\Rightarrow \ usin(\alpha - \beta)T - \frac{1}{2}\text{ }gcos\beta{\text{ }T}^{2} = 0 }{\Rightarrow \ T = \frac{2usin(\mathbf{\alpha} - \mathbf{\beta})}{gcos\mathbf{\beta}} }\](b) Range (R) Along the incline

\[R = \Delta x\text{~}\text{till}\text{~}t = T\]

i.e. \(\ R = u_{x}T + \frac{1}{2}a_{x}T^{2}\ \left( \right.\ \) Here \(\left. \ a_{x} = - gsin\beta \neq 0 \right)\)
Putting values of \(u_{x},a_{x}\& T\), we get

\[R = \frac{2u^{2}sin(\alpha - \beta)cos\alpha}{g\cos^{2}\beta}\]

(c) Maximum range

From above formula, \(R = \frac{u^{2}\lbrack sin(2\alpha - \beta) - sin\beta\rbrack}{g\cos^{2}\beta}\)
\(R\) is maximum when \(sin(2\alpha - \beta) = 1\)
i.e. \(2\alpha - \beta = 90^{\circ}\)
\[\Rightarrow \ \alpha = 45^{\circ} + \frac{\beta}{2} \]Also, \(\ R_{\max} = \frac{u^{2}(1 - sin\beta)}{g\cos^{2}\beta} = \frac{u^{2}(1 - sin\beta)}{g\left( 1 - \sin^{2}\beta \right)}\)
\[\therefore\ R_{\max} = \frac{u^{2}}{\text{ }g(1 + sin\beta)} \](d) Greatest distance from incline
\(S = \Delta y\) when \(\ v_{y} = 0\) i.e. when \(u_{y} + at = 0\)
i.e. when \(usin(\alpha - \beta) - gcos\beta t = 0\)

\[\Rightarrow \ t = \frac{usin(\alpha - \beta)}{gcos\beta}\]

Now, \(\ S = u_{y}t + \frac{1}{2}a_{y}t^{2}\)
Putting values of \(u_{y},a_{y}\) and \(t\), we get

\[S = \frac{u^{2}sin(\alpha - \beta)}{2\text{ }gcos\beta}\]

Special cases :

(1) If projected normally (i.e. perpendicular) to the plane, i.e. angle with plane \((\alpha - \beta) = 90^{\circ}\) i.e. \(\alpha = \left( 90^{\circ} + \beta \right)\)

(2) If projected horizontally
i.e. \(\alpha = 180^{\circ}\) and angle with the incline \(= \alpha - \beta = 180^{\circ} - \beta\)

(3) If the particle strikes normally to the plane i.e. at the moment of strike, \(V_{x} = 0\)
i.e. \(ucos(\alpha - \beta) - gsin\beta t = 0\)

Practice Exercise

Q. 1 A juggler throws a boll vertically upward and catches it after 6 sec. Determine
(i) the initial velocity of the ball.
(ii) the maximum height attained by the ball.
(iii) the position of ball at \(t = 2sec\).
(iv) the time at which ball is 20 m below the topmost point
Q. 2 A healthy yeoman standing at a distance of 7 m from a 11.8 m high building sees a kid slipping from the top floor. With what uniform speed should he run to catch the kid at his arms height \((1.8\text{ }m)\) ?
Q. 3 A balloon starts rising from the ground with an acceleration of \(1.25\text{ }m/s^{2}\). After 8 s a stone is released from the balloon. Find the time taken by the stone (after its fall) to reach ground.
Q. 4 A particle is projected in such a way that its position coordinates vary with time as \(y = 8t - 5t^{2}\) and \(x = 6t\) (taking point of projection as origin). What is the range of projectile?
(A) 48 m
(B) 4.8 m
(C) 9.6 m
(D) 24 m
Q. 5 A ball is projected upwards from the top of tower with a velocity \(50\text{ }m/s\) making an angle \(30^{\circ}\) with the horizontal. The height of the tower is 70 m . After how many seconds from the instant of throwing will the ball reach the ground ?
(A) 2 s
(B) 5 s
(C) 7 s
(D) 9 s
Q. 6 A particle is projected upwards with a velocity of \(100\text{ }m/s\) at an angle of \(60^{\circ}\), with the vertical. Then time taken by the particle when it will move perpendicular to its initial direction-
(A) 10 sec
(B) \(\frac{20}{\sqrt{3}}\sec\)
(C) 5 sec
(D) \(10\sqrt{3}\sec\)
Q. 7 A particle is projected at an angle ' \(\alpha\) ' to the horizontal. Up and down there is a plane in case (a) & case (b), inclined at an angle \(\beta\) to the horizontal. If the ratio of time of flights on these plane in case (a) & case (b) be \(1:2\), then the ratio \(\frac{tan\alpha}{tan\beta}\) is equal to:

(A) \(\frac{2}{1}\)
(B) \(\frac{3}{1}\)
(C) \(\frac{4}{1}\)
(D) \(\frac{5}{3}\)

Paragraph for question no. 8 to 10

An inclined plane makes an angle \(\theta = 45^{\circ}\) with horizontal. A stone is projected normally from the inclined plane, with speed \(u\text{ }m/s\) at \(t = 0x\) and \(y\) axis are drawn from point of projection along and normal to inclined plane as shown. The length of incline is sufficient for stone to land on it and neglect air friction.

Q. 8 The instant of time at which velocity of stone is parallel to x -axis
(A) \(\frac{2\sqrt{2}u}{g}\)
(B) \(\frac{2u}{g}\)
(C) \(\frac{\sqrt{2}u}{g}\)
(D) \(\frac{u}{\sqrt{2}\text{ }g}\)
Q. 9 The instant of time at which velocity of stone makes an angle \(\theta = 45^{\circ}\) with positive x -axis
(A) \(\frac{2\sqrt{2}u}{g}\)
(B) \(\frac{2u}{g}\)
(C) \(\frac{\sqrt{2}u}{g}\)
(D) \(\frac{u}{\sqrt{2}\text{ }g}\)
Q. 10 The instant of time till which (starting from \(t = 0\) ) component of displacement along x -axis is half the range on inclined plane is
(A) \(\frac{2\sqrt{2}u}{g}\)
(B) \(\frac{2u}{g}\)
(C) \(\frac{\sqrt{2}u}{g}\)
(D) \(\frac{u}{\sqrt{2}\text{ }g}\)

Answers

Q. 1 (i) \(30\text{ }m/s\) (ii) 45 m (iii) 40 m (iv) 1 sec . and 5 sec
Q. \(2\frac{7}{\sqrt{2}}\text{ }m/s\ \) Q. \(3\ 4sec\ \) Q. \(4\ \) (C) 9.6 m
\[{\begin{matrix} \text{~}\text{Q.}\text{~}5 & \text{~}\text{(C)}\text{~}7\text{ }s & \text{~}\text{Q.}\text{~}6\text{~}\text{(B)}\text{~}\frac{20}{\sqrt{3}}\sec \end{matrix} }{\begin{matrix} \text{~}\text{Q.}\text{~}7 & \text{~}\text{(B)}\text{~}\frac{3}{1} & \text{~}\text{Q.}\text{~}8 & \text{~}\text{(C)}\text{~}\frac{\sqrt{2}u}{g} \end{matrix} }\]Q. 9 (D) \(\frac{u}{\sqrt{2}\text{ }g}\ \) Q. \(10\ \) (B) \(\frac{2u}{g}\)
Q. 8 The instant of time at which velocity of stone is parallel to x -axis
(A) \(\frac{2\sqrt{2}u}{g}\)
(B) \(\frac{2u}{g}\)
(C) \(\frac{\sqrt{2}u}{g}\)
(D) \(\frac{u}{\sqrt{2}\text{ }g}\)
Q. 9 The instant of time at which velocity of stone makes an angle \(\theta = 45^{\circ}\) with positive x -axis
(A) \(\frac{2\sqrt{2}u}{g}\)
(B) \(\frac{2u}{g}\)
(C) \(\frac{\sqrt{2}u}{g}\)
(D) \(\frac{u}{\sqrt{2}\text{ }g}\)
Q. 10 The instant of time till which (starting from \(t = 0\) ) component of displacement along x -axis is half the range on inclined plane is
(A) \(\frac{2\sqrt{2}u}{g}\)
(B) \(\frac{2u}{g}\)
(C) \(\frac{\sqrt{2}u}{g}\)
(D) \(\frac{u}{\sqrt{2}\text{ }g}\)

Answers

Q. 1 (i) \(30\text{ }m/s\) (ii) 45 m (iii) 40 m (iv) 1 sec . and 5 sec
Q. \(2\frac{7}{\sqrt{2}}\text{ }m/s\ \) Q. \(3\ 4sec\ \) Q. \(4\ \) (C) 9.6 m
\[{\begin{matrix} \text{~}\text{Q.}\text{~}5 & \text{~}\text{(C)}\text{~}7\text{ }s & \text{~}\text{Q.}\text{~}6\text{~}\text{(B)}\text{~}\frac{20}{\sqrt{3}}\sec \end{matrix} }{\begin{matrix} \text{~}\text{Q.}\text{~}7 & \text{~}\text{(B)}\text{~}\frac{3}{1} & \text{~}\text{Q.}\text{~}8 & \text{~}\text{(C)}\text{~}\frac{\sqrt{2}u}{g} \end{matrix} }\]Q. 9 (D) \(\frac{u}{\sqrt{2}\text{ }g}\ \) Q. \(10\ \) (B) \(\frac{2u}{g}\)

Solved Example

Q. 1 The motion of an object falling from rest in a viscous medium can be described by the equation \(a = \alpha - \beta v\). Where a and v are the acceleration and velocity of the object and \(\alpha\) and \(\beta\) are constants. Find.
(i) the initial acceleration.
(ii) the velocity at which acceleration becomes zero.
(iii) the velocity as a function of time.

Sol. (i) The initial velocity of object \(v = 0\) So, initial acceleration a \(= \alpha - (\beta \times 0) = \alpha\)
(ii) For acceleration to be zero

\[0 = \alpha - \beta v\ \text{~}\text{or,}\text{~}\ v = \frac{\alpha}{\beta}\]

Note : The velocity at which acceleration reduces to zero is the maximum velocity with which an object will fall in a viscous medium. This velocity is called Terminal velocity.
(iii) \(a = \frac{dv}{dt}\ \) or, \(\ \alpha - \beta v = \frac{dv}{dt}\ \) or, \(\frac{dv}{\alpha - \beta v} = dt\)

Integrating the expression with boundary conditions : \(t = 0,v = 0\) and \(t = t,v = v\)
\(\int_{0}^{v}\mspace{2mu}\frac{dv}{\alpha - \beta v} = \int_{0}^{t}\mspace{2mu} dt\ \) or \(\left( \frac{- 1}{\beta} \right)ln(\alpha - \beta v)b_{b}^{v} = tb_{b}^{t}\)
\[\Rightarrow \ - \frac{1}{c}\lbrack\mathcal{l}n(\alpha - \beta v) - \mathcal{l}n(\alpha)\rbrack = t - 0\ \Rightarrow ln\left( \frac{\alpha - \beta v}{\alpha} \right) = - \beta t \]Rearranging the terms, we get

\[v = \frac{\alpha}{\beta}\left( 1 - e^{- \beta t} \right)\]

Q. 2 At \(t = 0\), a particle is at rest at origin. Its acceleration is \(2\text{ }m/s^{2}\) for first 2 sec . and \(- 2\text{ }m/s^{2}\) for next 4 sec as shown in a versus t graph.

Plot graphs for
(i) Velocity versus time
(ii) speed versus time
(iii) Displacement versus time
(iv) Distance versus time

Sol. (i) \(\ V_{2} - V_{0} =\) Area of a Vs t graph for \(t = 0\) to \(t = 2sec\)

\[V_{2} - 0 = 2 \times 2 \Rightarrow \ V_{2} = + 4\text{ }m/s\]

Now \(\ V_{6} - V_{2} = - 2 \times 4 \Rightarrow \ {\text{ }V}_{6} = - 4\text{ }m/s\)

(ii) Since slope of a Vs \(t\) graph from \(t = 2\) to 6 sec . is constant, we can observe its speed i.e. magnitude of its velocity is zero at \(= 4\sec\). and after that magnitude of velocity increases in negative direction up to \(4\text{ }m/s\) at the same rate.
(iii) Displacement (x) Vst
\(x_{2} - x_{0} =\) area of \(v\) vst graph for \(t = 0,t = 2sec\)
\[{x_{2} - 0 = \frac{1}{2}(2)(4) \Rightarrow x_{2} = + 4\text{ }m }{x_{4} - x_{2} = \frac{1}{2}(4)(2) }\]

\[x_{4} = 8\text{ }m \]also \(x_{6} - x_{4} = \frac{l}{2}( - 4)(2) = - 4\text{ }m\)
\[\therefore\ x_{6} = + 4\text{ }m \](iv) Distance (d) vs t
\[{d_{2} - d_{0} = \frac{1}{2}(2)(4)\ \Rightarrow \ d_{2} = 4\text{ }m }{d_{4} - d_{2} = \frac{1}{2}(2)(4)\ \Rightarrow \ d_{4} = 8\text{ }m }\]Also \(d_{6} - d_{4} = \left| \frac{I}{2}(2)( - 4) \right| = 4\)

\[\Rightarrow d_{6} = 12\text{ }m \]Q. 3 Three particles are projected from same point and their paths are as shown. Compare their horizontal and vertical component of velocities of projection
Sol.

\[\begin{matrix} & & H_{A} = H_{B} > H_{C} \Rightarrow \left( u_{y} \right)_{A} = \left( u_{y} \right)_{B} > \left( u_{y} \right)_{C} \\ \therefore & & R_{B} > R_{A} \\ & \text{~}\text{i.e.}\text{~} & \left( u_{x}u_{y} \right)_{B} > \left( u_{x}u_{y} \right)_{A}\ \left\lbrack \because\text{~}\text{their}\text{~}u_{x}\text{~}\text{are equal}\text{~} \right\rbrack \\ \therefore & & \left( u_{x} \right)_{B} > \left( u_{x} \right)_{A} \\ \text{~}\text{Also}\text{~} & & R_{B} = R_{C} \\ & \text{~}\text{i.e.}\text{~} & \left( u_{x}u_{y} \right)_{B} = \left( u_{x}u_{y} \right)_{C} \\ & \text{~}\text{but}\text{~} & \left( u_{y} \right)_{B} > \left( u_{y} \right)_{C} \\ & \therefore & \left( u_{x} \right)_{C} > \left( u_{x} \right)_{B} > \left( u_{x} \right)_{C} \end{matrix}\]

Q. 4 A car goes out of control and slides off a steep embankment of height \(h\) at \(\theta\) to the horizontal. It lands in a ditch at a distance \(R\) from the base. Find the speed at which the car leaves the slope. (Take \(h = 12.5\text{ }m;R = 10\text{ }m;\theta = 45^{\circ}\) )

Sol.

\[\begin{matrix} \Delta x = 10 = \left( ucos45^{\circ} \right)t \Rightarrow t = \frac{10\sqrt{2}}{u} \\ \Delta y = - \left( usin45^{\circ} \right)t - \frac{1}{2}{gt}^{2} = - 12.5 \\ \Rightarrow \ \left( \frac{u}{\sqrt{2}} \times \frac{10\sqrt{2}}{u} \right) + \frac{1}{2}(10)\left( \frac{10\sqrt{2}}{u} \right)^{2} = 12.5 \end{matrix}\]

On solving we get \(u = 20\text{ }m/s\)
Q. 5 Find range of projectile on the inclined plane which is projected perpendicular to the incline plane with velocity \(20\text{ }m/s\) as shown in figure.
Sol. \(\ \beta = 37^{\circ}\)

\[\alpha - \beta = 90^{\circ}\&\alpha = 90^{\circ} + \beta = 90^{\circ} + 37^{\circ}\]

\[\begin{matrix} \therefore & \text{~}\text{Range,}\text{~}R = \frac{2(20)^{2}sin\left( 90^{\circ} \right)cos\left( 90^{\circ} + 37^{\circ} \right)}{10 \times \cos^{2}\left( 37^{\circ} \right)} \\ & = \frac{2(400)}{10(4/5)^{2}} \times \left( - \frac{3}{5} \right)\ \left\lbrack \because cos\left( 90^{\circ} + \theta \right) = - sin\theta \right\rbrack \\ \therefore & R = - 75\text{ }m \\ \therefore & |R| = 75\text{ }m \end{matrix}\]

Here negative sign shown that particle strikes the plane along down the incline

Relative Motion

Relative Velocity

It is given by the time rate of change of position of one object w.r.t. another. Relative velocity of a body \(B\) with respect to some other body A means velocity of B is recorded by an observer sitting on A . Mathematically.

Relative velocity of B w.r.t. A : \({\overrightarrow{v}}_{B/A} = {\overrightarrow{v}}_{B} - {\overrightarrow{v}}_{A}\)

Proof: \({\overrightarrow{r}}_{B/A} = {\overrightarrow{r}}_{B} - {\overrightarrow{r}}_{A}\). Differentiation this equation w.r.t. time, we get

\[\frac{d\left( {\overrightarrow{r}}_{B/A} \right)}{dt} = \frac{d{\overrightarrow{r}}_{B}}{dt} - \frac{d{\overrightarrow{r}}_{A}}{dt}\ \text{~}\text{but}\text{~}\ \frac{d{\overrightarrow{r}}_{A}}{dt} = {\overrightarrow{v}}_{A},\frac{\text{ }d{\overrightarrow{r}}_{B}}{dt} = {\overrightarrow{v}}_{B}\text{~}\text{and}\text{~}\frac{d\left( {\overrightarrow{r}}_{B/A} \right)}{dt} = {\overrightarrow{v}}_{B/A}\]

putting these values we get \({\overrightarrow{v}}_{B/A} = {\overrightarrow{v}}_{B} - {\overrightarrow{v}}_{A}\). Hence proved.
Similarly, we can prove that relative velocity of A w.r.t. B :
\[{\overrightarrow{v}}_{A/B} = {\overrightarrow{v}}_{A} - {\overrightarrow{v}}_{B}\]

Graphical Method to find Relative Velocity

When two bodies move at angle \(\theta\) with each other then their relative velocity is given by :
Magnitude : \(\left| {\overrightarrow{v}}_{B/A} \right| = \left| {\overrightarrow{v}}_{B} - {\overrightarrow{v}}_{A} \right| = \sqrt{v_{A}^{2} + v_{B}^{2} + 2v_{A}v_{B}cos(180 - \theta)} = \sqrt{v_{A}^{2} + v_{B}^{2} - 2v_{A}v_{B}cos\theta}\)
Direction : \(tan\alpha = \frac{v_{B}sin(180 - \theta)}{v_{A} + v_{B}cos(180 - \theta)}\)
\[\Rightarrow tan\alpha = \frac{v_{B}sin\theta}{v_{A} - v_{B}cos\theta} \]If \(\theta = 90^{\circ}\) then \(\left| {\overrightarrow{v}}_{B/A} \right| = \sqrt{v_{A}^{2} + v_{B}^{2}}\) and \(tan\alpha = \frac{v_{B}}{v_{A}}\)

We can find the velocity of a particle in a frame if we know the particle's velocity in some other frame and the relative velocity of frames w.r.t each other.
In two observers are watching a moving particle P from the origins of reference A and B , while B moves at a constant velocity \({\overrightarrow{v}}_{B/A}\) relative to A .

Fig. shows a certain instant during the motion. At this instant, the position vector of \(B\) relative to \(A\) is \({\overrightarrow{r}}_{BA}\). Also, the position vectors of particle P are \({\overrightarrow{r}}_{P/A}\) relative to A and \({\overrightarrow{r}}_{P/B}\) relative to B . From the arrangement of heads and tails of those three position vectors, we can relate the vectors with \({\overrightarrow{r}}_{P/A} = {\overrightarrow{r}}_{P/B} + {\overrightarrow{r}}_{B/A}\). By taking the time derivative of this equation, we can relate the velocities \({\overrightarrow{v}}_{P/A} = {\overrightarrow{v}}_{P/B} + {\overrightarrow{v}}_{B/A}\). We can understand the concept of relative velocity by a simple situation as follows :

Rain - Man Problems

Formula to be applied: \({\overrightarrow{v}}_{r/m} = {\overrightarrow{v}}_{r} - {\overrightarrow{v}}_{m}\), where \({\overrightarrow{v}}_{r/m}\) is velocity of rain w.r.t. man, \({\overrightarrow{v}}_{r}\) is the velocity of rain (w.r.t. ground), and \({\overrightarrow{v}}_{m}\) is the velocity of man (w.r.t. ground).

If rain is falling vertically downwards with a speed \(v_{r}\) and a man is running horizontally towards east with a speed \(v_{m}\).

What is the relative velocity of rain w.r.t. man?

Given: \({\overrightarrow{v}}_{r} = - v_{r}\widehat{j},{\overrightarrow{v}}_{m} = v_{m}\widehat{i}\),
Now \({\overrightarrow{v}}_{r/m} = {\overrightarrow{v}}_{r} - {\overrightarrow{v}}_{m} = - v_{r}\widehat{j} - v_{m}\widehat{i} \Rightarrow {\overrightarrow{v}}_{r/m} = - v_{m}\widehat{i} - v_{r}\widehat{j}\).
Magnitude: \(\sqrt{v_{m}^{2} + v_{r}^{2}}\) and direction: \(tan\alpha = \frac{v_{m}}{v_{r}}\)

Example : If rain is already falling at some angle \(\theta\) with horizontal, then with what velocity the man should travel so that the rain appears vertically downwards to him?

Here, \({\overrightarrow{v}}_{m} = v_{m}\widehat{i},{\overrightarrow{v}}_{r} = v_{r}sin\theta\widehat{i} - v_{r}cos\theta\widehat{j}\)
Now, \({\overrightarrow{v}}_{r/m} = {\overrightarrow{v}}_{r} - {\overrightarrow{v}}_{m} = \left( v_{r}\sin\theta - v_{m} \right)\widehat{i} - v_{r}cos\theta\widehat{j}\)
Now for rain to appear falling vertically, the horizontal component of \({\overrightarrow{v}}_{r/m}\) should be zero, i.e.,

\[\begin{matrix} & v_{r}sin\theta - v_{m} = 0 \Rightarrow sin\theta = \frac{v_{m}}{v_{r}}\text{~}\text{and}\text{~}\left| v_{r/m} \right| = v_{r}cos\theta \\ & \ = v_{r}\sqrt{1 - \sin^{2}\theta} = v_{r}\sqrt{1 - \frac{v_{m}^{2}}{v_{r}^{2}}} \\ & \text{~}\text{or}\text{~}v_{r/m} = \sqrt{v_{r}^{2} - v_{m}^{2}} \end{matrix}\]

We can illustrate the whole situation by the diagrams.
It is quite interesting to notice the steady rainfall sitting in a vehicle such as bus, car, etc. While moving on a straight track the direction of rainfall changes when the vehicle changes its velocity. That means, the velocity of the rain you observe is the velocity of the rain relative to you. Therefore, your observed velocity of rainfall (both magnitude and direction of velocity of rainfall) is the velocity of the rain with respect to the vehicle (you). If you measure the velocity of the rainfall while the vehicle is stationary, that gives actual velocity of rainfall.

Remember following points regarding relative motion :

• If the velocity is mentioned without specifying the frame, assume it is with respect to the ground.

• In many cases, a body travels on water or in air. Depending on the context you will have to figure out whether the velocity is with respect to the water/air or with respect to the ground.

Let us analyse following situation.
The man is stationary and the rain is falling at his back to an angle \(\phi\) with the vertical

\[\begin{matrix} v_{rm} & \ = v_{r}\ v_{m} = 0 \\ & \theta = \phi\ \text{~}\text{here}\text{~}\theta = \text{~}\text{Angle at which rain appears to man}\text{~} \end{matrix}\]

Now man starts moving forward with speed \(v_{m}\). The relative velocity of rain w.r.t. man shifts towards vertical direction.

As the man further increase his speed, then at a particular value the rain appears to be falling vertically.

Velocity vector diagram

If the man increases his speed further more, then rain appear to be falling from the forward direction.

Notice in above figure how man changes orientation of umbrella to prevent himself from rain

River-Swimmer Problems

When a man or a boat is swimming in water, he generates a velocity relative to water ( \(vm/s\) ) by his own efforts. Actual velocity of man in water will be a resultant of man's effort and the river velocity (u m/s). Down stream : Man makes efforts in direction of flow, the velocity of man w.r.t. ground is \((u + v)m/s\) as shown below.

Up stream: Man makes efforts opposite to the direction of flow, the velocity of man w.r.t. ground is \((v - u)m/s\) as shown below.

Note following points :

(i) Swimmer keeps himself at an angle of \(30^{\circ}\) with river flow mean the velocity of swimmer is w.r.t river flow.
(ii) A man swims in water \(\Rightarrow\) velocity of man w.r.t. water.
(iii) A swimmer heads to means (velocity is not w.r.t. ground)
(iv) Person wants to go to destination then direction of velocity is w.r.t. ground.

Let discuss a situation when swimmer & river velocity are known
Suppose velocity of river is \(u\) and swimmer can swim with a velocity ' \(v\) ' w.r.t. river flow.
(a) What should be the angle \(\theta\) with the river flow at which the man should swim so that the time taken to cross the river be minimum?

Sol. Let man storts swimming at an angle as shown in figure.

\[\begin{matrix} & {\overrightarrow{v}}_{m} = {\overrightarrow{v}}_{m} + {\overrightarrow{v}}_{r} \\ & \ = ( - Vsin\theta\widehat{i} + vcos\theta\widehat{j}) + u\widehat{i} \\ & \ = (u - vsin\theta)\widehat{i} + (vcos\theta)\widehat{j} \\ & \ = \text{~}\text{If width of river is 'd' then t}\text{ime to cross.}\text{~} \\ & t = \frac{d}{vcos\theta} \end{matrix}\]

for \(t_{\text{min}\text{~}},cos\theta = 1\) at \(\theta = 0^{\circ}\)

\[t_{\min} = \frac{d}{v}\]

So the man should try to swim perpendicular to the river flow to minimize the time in each case.
(b) What should be the angle \(\theta\) at which the man should swim so that the length of path be minimum? for minimum length of the path, drift \(x\) should be minimum.

Sol. Drift for given situation \(=\) time \(\times \left\{ {\overrightarrow{v}}_{m} \right.\ \) along the flow \(\}\)

\[\begin{matrix} & & x = \frac{d}{vcos\theta} \times (u - vsin\theta) \\ & x = \frac{du}{v}sec\theta - dtan\theta & \text{(A)} \end{matrix}\]

Case-I

\(v > u\) or the river flow is less than the velocity of man's effort.
In such case the minimum possible drift will be zero. So the man should has swim at the angle.

\[\begin{matrix} & x = 0 \Rightarrow u - vsin\theta = 0 \\ & sin\theta = \frac{u}{v} \end{matrix}\]

Case-II

\(v < u\) or the river flow is greater than velocity of man's effort.
In such a case, boat cannot reach the point directly opposite to its starting point. i.e. drift can never be zero.
\(\therefore\) for x to be minimum

\[\frac{dx}{\text{ }d\theta} = 0\]

Differentialing equation (A)

\[\begin{matrix} \therefore & \frac{dx}{\text{ }d\theta} = \frac{du}{v}sec\theta tan\theta - {dsec}^{2}\theta = 0 \\ \therefore & sin\theta = \frac{v}{u} \end{matrix}\ \theta = \sin^{- 1}\frac{v}{u}\]

Thus, to minimize the drift, boat starts at an angle \(\theta\) from the river flow.

In this case minimum drift can be calculated by putting value of

\[\begin{matrix} & \theta = \sin^{- 1}\left( \frac{v}{u} \right)\text{~}\text{in equation}\text{~}(A) \\ & x_{\min} = \frac{d\sqrt{u^{2} - v^{2}}}{v} \end{matrix}\]

Practice Exercise

Q. 1 A man is trying to cross the river 100 m wide by a boat. The river is flowing with the velocity \(5\text{ }m/s\) and the boat's velocity in still water is \(3\text{ }m/s\). Find the minimum time in which he can cross the river and the drift in this case?
Q. 2 Find the direction in which the man (of above illstration) should row so as to have minimum drift. Also find the minimum possible drift and the time taken to cross the river in this case?

Answers

Q. \(1\ \frac{100}{3}sec,\frac{500}{3}\text{ }m\)
Q. \(2sin\theta = \frac{3}{5},x_{\min} = \frac{400}{3}\text{ }m,t = \frac{125}{3}\sec\).

Swimming in a desired direction:

Many times the person is not interested in minimizing the time or drift. But he has to reach a particular place. This is common in the cases of an airplane or motor boat.

The man desires to have this final velocity along \(AB\) in other words he has to move from \(A\) to \(B\). We wish to find the direction in which he should make an effort so that his actual velocity is along line \(AB\), w.r.t. ground. In this method we assume AB to be the reference line the resultant of v and u is along line AB . Thus the components of \(v\) and \(u\) in a direction perpendicular to line \(AB\) should cancel each other.

So v \(sin\alpha = usin\theta\)
or \(sin\alpha = \frac{usin\theta}{v}\)
here \(\theta\), \(u\& v\) are given in a problem, so we can calculate \(\alpha\) by above relation

Closest distance of approach between two bodies

Relative Motion Between two Projectiles

Let us now discuss the relative motion between two projectiles or the path observed by one projectile of the other. Suppose that two particles are projected from the ground with speeds \(u_{1}\) and \(u_{2}\) at angles \(\alpha_{1}\) and \(\alpha_{2}\) as shown in figure. Acceleration of both the particles is \(g\) downwards. So, relative acceleration between them is zero because
\(a_{12} = a_{1} - a_{2} = g - g =\) zero

i.e., the relative motion between the two particles is uniform. Now

Therefore, and

\[\begin{matrix} u_{1x} = u_{1}cos\alpha_{1}, & u_{2x} = u_{2}cos\alpha_{2} \\ u_{1y} = u_{1}sin\alpha_{1},\text{~}\text{and}\text{~} & u_{2y} = u_{2}sin\alpha_{2} \end{matrix}\]

\(u_{12x}\) and \(u_{12y}\) are the \(x\) and \(y\) components of relative velocity of 1 with respect to 2 .
Hence, relative motion of 1 with respect to 2 is a straigh line at an angle \(\theta = \tan^{- 1}\left( \frac{u_{12y}}{u_{12x}} \right)\) with positive x -axis.
Now, if \(u_{12x} = 0\) or \(u_{1}cos\alpha_{1} = u_{2}cos\alpha_{2}\), the relative motion is along \(y\)-axis or in vertical direction (as \(\theta = 90^{\circ}\) ). similarly, if \(u_{12y} = 0\) or \(u_{1}sin\alpha_{1} = u_{2}sin\alpha_{2}\), the relative motion is along \(x\) -

axis or in horizontal direction (as \(\theta = 0^{\circ}\) ).

Condition of Collision of two Projectiles

From the above discussion, it is clear that relative motion between two projectiles is uniform and the path of one projectile as observed by the other is a straight line. Now let the particles are projected simultaneously from two different heights \(h_{1}\) and \(h_{2}\) with speeds \(u_{1}\) and \(u_{2}\) in the directions shown in figure. Then the particles collide in air if relative velocity of 1 with respect to \(2\left( {\overrightarrow{u}}_{12} \right)\) is along line AB or the relative velocity of 2 with respect to \(1\left( {\overrightarrow{u}}_{12} \right)\) is along the line BA . Thus,

\[tan\theta = \frac{u_{12y}}{u_{12x}} = \left( \frac{h_{2} - h_{1}}{\text{ }s} \right)\]

Here

\[u_{12y} = u_{1}sin\alpha_{1} - u_{2}sina_{2}\]

and \(\ u_{12x} = \left( u_{1}cos\alpha_{1} \right) - \left( - u_{2}cos\alpha_{2} \right) = u_{1}cos\alpha_{1} + u_{2}cos\alpha_{2}\)
If both the particles are initially at the same level ( \(\left. \ h_{1} = h_{2} \right)\), then for collision

\[u_{12y} = 0\ \text{~}\text{or}\text{~}\ u_{1}sin\alpha_{1} = u_{2}sin\alpha_{2}\]

The time of collision of the two particles will be

\[t = \frac{AB}{\left| {\overrightarrow{u}}_{12} \right|} = \frac{AB}{\sqrt{\left( u_{12x} \right)^{2} + \left( u_{12y} \right)^{2}}}\]

Further, the above conditions are not merely sufficient for collision to takes place. For example, the time of collision discussed above should be less than the time of collision of either of the particles with the ground.

Solved Example

Q. 1 Ram crossing a 2.5 m wide conveyor belt moves with a speed of \(1.6\text{ }m/s\). The conveyor belt moves at uniform speed of \(1.2\text{ }m/s\).
(A) If the Ram walks straight across the belt, determine the velocity of the Ram relative to an observer standing on ground.
Sol. If you walk across a conveyor belt while the conveyor belt takes you along the length, you will not be able to move directly across the conveyor belt, but will end up down the length.
Here the velocity of the Ram will be net effect of his own motion and due to motion of conveyor belt
The velocity of the Ram relative to the conveyor belt \(v_{rf}\), is same as velocity of Ram if conveyor belt was still,
\(v_{c}\) is the velocity of the conveyor belt
we need to find \(\mathbf{v}_{\mathbf{r}}\), the velocity of the Ram relative to the Earth.
Writing Equation of net motion \(\mathbf{v}_{\mathbf{r}} = \mathbf{v}_{\mathbf{rc}} + \mathbf{v}_{\mathbf{c}}\).
three vectors are shown in Figure (a). The quantity \(\mathbf{v}_{\mathbf{rc}}\) is due \(y;\mathbf{v}_{\mathbf{c}}\) is due \(x\); and the vector sum of the two, \(\mathbf{v}_{\mathbf{r}}\), is at an angle \(\theta\) as defined in Figure (a).
(a)

the speed \(v_{r}\) of the Ram relative to the Earth is

\[v_{r} = \sqrt{v_{r/c}^{2} + v_{c}^{2}}\]

(B) If Shyam has same speed on a still conveyor belt, and is to reach directly across the same moving conveyor belt. At what angle should he walk?
(b)

Sol. To go straight across the conveyor belt he has to walk at some angle.
Writing Equation of net motion \(\mathbf{v}_{s} = \mathbf{v}_{s/c} + \mathbf{v}_{c}\).
three vectors are shown in Figure (b)
As in part (b), we know \(\mathbf{v}_{\mathbf{c}}\) and the magnitude of the vector \(\mathbf{v}_{\mathbf{s}/\mathbf{c}}\), and we want \(\mathbf{v}_{\mathbf{c}}\) to be directed across the conveyor belt.

\[v_{s} = \sqrt{v_{s/c}^{2} - v_{c}^{2}}\]

Note the difference between the triangle in Figure (a) and the one in Figure (b)
Q. 2 An aeroplane pilot wishes to fly due west. A wind of \(100\text{ }km/h\) is blowing toward the south
(A) What is the speed of the plane with respect to ground ?
(B) If the airspeed of the plane (its speed in still air) is \(300\text{ }km/h\), in which direction should the pilot head?

Sol.
(A) Given,

Velocity of air with respect of ground \({\overrightarrow{v}}_{\text{A/G}\text{~}} = 100\text{ }km/hr\)
Velocity of plane with respect to air \({\overrightarrow{v}}_{\text{P}\text{/A}\text{~}} = 300\text{ }km/hr\)

(B) As the plane is to move towards west, due to air in south direction, air will try drift the plane in south direction., air will try to drift the plane in south direction. Hence, the plane has to make an angle \(\theta\) towards north-west, south west direction, in order to reach at point on west.

\[{\overrightarrow{v}}_{P/A} = {\overrightarrow{v}}_{P/G} - {\overrightarrow{v}}_{A/G}\text{~}\text{and}\text{~}V_{P/A}sin\theta = V_{AG}\]

Q. 3 Ariver flows due south with a speed of \(2.0\text{ }m/s\). A man steers a motorboat across the diver: his velocity relative to the water is \(4\text{ }m/s\) due east. The river is 800 m wide.
(A) What is his velocity (magnitude direction) relative to the earth?
(B) How much time is required to cross the river?
(C) How far south of his starting point will be reach the opposite bank?

Sol. Velocity of river (i.e., speed of river w.r.t. earth) \({\overrightarrow{v}}_{re} = 2\text{ }m/s\)
Width of the river \(= 800\text{ }m\)

According to the given statement the diagram will be as given
(A) When two vectors are acting at an angle of \(90^{\circ}\), their resultant can be obtained by pythagorous theorem,

\[{\overrightarrow{v}}_{be} = \sqrt{v_{br}^{2} + v_{re}^{2}} = \sqrt{16 + 4} = \sqrt{20} = 4.6\text{ }m/s\]

To find direction, we have

\[tan\theta = \frac{v_{re}}{v_{br}} = \frac{2}{4} = \frac{1}{2} \Rightarrow \ \theta = \tan^{- 1}\left( \frac{1}{2} \right)\]

(B) Time taken to cross the river \(= \frac{\text{~}\text{Displacement of boat }\text{w.r.}\text{t.river}\text{~}}{\text{~}\text{Velocity of boat w.r.t river}\text{~}}\)

\[\Rightarrow \frac{800}{4} = 200\text{ }s\]

(C) Desired position on other side is A , but due to current of river boat is drifted to position B . To find out this drift we need time taken in all to cross the river (200s) and speed of current ( \(2{\text{ }ms}^{- 1}\) )
So the distance \(AB =\) Time taken \(\times\) speed of current \(= 200 \times 2 = 400\text{ }m\)
Hence, the boat is drifted by 400 m away from position A .
Q. 4 A person walks up a stationary escalator in \(t_{1}\) second. If he remains stationary on the escalator, then it can take him up in \(t_{2}\) second. If the length of the escalator is L , then
(A) Determine the speed of man with respect to the escalator.
(B) Determine the speed of the escalator.
(C) How much time would it take him to walk up the moving escalator?

Sol.
(A) As the escalator is stationary, so the distance covered in \(t_{1}\) second is L which is the length of the escalator.

\[\text{~}\text{Speed of the man }\text{w.r.t.}\text{ the escalator}\text{~}v_{me} = \frac{L}{t_{1}}\]

(B) When the man is stationary, by taking man as reference point the distance covered by the escalator is L in time \(t_{2}\).

\[\text{~}\text{Speed of escalator}\text{~}v_{e} = \frac{L}{t_{2}}\]

(C) Speed of man w.r.t. the ground

\[\begin{matrix} & v_{m} = v_{me} + v_{e} \\ \Rightarrow \ & v_{m} = \frac{L}{t_{1}} + \frac{L}{t_{2}} = L\left\lbrack \frac{1}{t_{2}} + \frac{1}{t_{2}} \right\rbrack = L\left\lbrack \frac{t_{1} + t_{2}}{t_{1}t_{2}} \right\rbrack \\ \Rightarrow \ & L = v_{m}\left\lbrack \frac{t_{1}t_{2}}{t_{1} + t_{2}} \right\rbrack \\ & \left\lbrack \frac{t_{1}t_{2}}{t_{1} + t_{2}} \right\rbrack\text{~}\text{is the time taken by the man to walk up the moving escalator.}\text{~} \end{matrix}\]

Q. 5 A person standing on a road has to hold his umbrella at \(60^{\circ}\) with the vertical to keep the rain away. He throws the umbrella and starts running at \(20{\text{ }ms}^{- 1}\). He finds that rain drops are falling on him vertically. Find the speed of the rain drops with respect to
(A) the road, and
(B) the moving person.

Sol. Given \(\theta = 60^{\circ}\) and velocity person
\({\overrightarrow{v}}_{p} = \overrightarrow{OA} = 20{\text{ }ms}^{- 1}\).
This velocity is the same as the velocity of person w.r.t ground. First of all let's see how the diagram works out.
\({\overrightarrow{v}}_{rP} = \overrightarrow{OB} =\) velocity of rain w.r.t. the person.

\({\overrightarrow{v}}_{r} = \overrightarrow{OC} =\) velocity of rain w.r.t. earth \({\overrightarrow{v}}_{rP}\) is along \(\overrightarrow{OB}\) as a person has to hold umbrella at an angle with vertical which is the angle between velocity of rain and velocity of rain w.r.t. the person.

Values of \({\overrightarrow{v}}_{r}\) and \({\overrightarrow{v}}_{rp}\) can be obtained by using simple trigonometric relations.
(A) Speed of rain drops w.r.t. earth \(= {\overrightarrow{v}}_{r} = \overrightarrow{OC}\)

From \(\bigtriangleup OCM,\frac{CB}{OC} = sin60^{\circ} \Rightarrow \ OC = \frac{CB}{sin60^{\circ}}\)

\[= \frac{20}{\sqrt{3}/2} = \frac{40}{\sqrt{3}} = \frac{40\sqrt{3}}{3}{\text{ }ms}^{- 1}\]

(B) Speed of rain w.r.t. the person \({\overrightarrow{v}}_{rP} = \overrightarrow{OB}\)

From \(\bigtriangleup OCM,\frac{OB}{CB} = cot60^{\circ}\)
\[\Rightarrow \ OB = CBcot60^{\circ} = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3}{\text{ }ms}^{- 1} \]Q. 6 A large heavy box is sliding without friction down a smooth plane of inclination \(\theta\). From a point P on the bottom of a box, a particle is projected inside the box. The initial speed of the particle with respect to box is u and the direction of projection makes an angle \(\alpha\) with the bottom as
shown in figure.

(a) Find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands. (Assume that the particle does not hit any other surface of the box. Neglect air resistance).
(b) If the horizontal displacement of the particle as seen by an observer on the ground is zero, find the speed of the box with respect to the ground at the instant when the particle was projected.
Sol.
(a) \(\ u\) is the relative velocity of the particle with respect ot the box. Resolve \(u\).
\(u_{x}\) is the relative velocity of particle with respect to the box in x - direction.
\(u_{y}\) is the relative velocity with respect to the box in y - direction.
Since, there is no velocity of the box in the y -direction, therefore this is the vertical velocity of the particle with respect to ground also.
Y - direction motion ( Taking relative terms w.r.t. box)

\[\begin{matrix} & u_{y} = + usin\alpha \\ & a_{y} = - gcos\theta \\ & {\text{ }s}_{y} = 0 \\ & {\text{ }s}_{y} = u_{y}t + \frac{1}{2}a_{y}t^{2} \Rightarrow 0 = (usin\alpha)t - \frac{1}{2}\text{ }gcos\theta \times t^{2} \\ & \ \Rightarrow t = \frac{2usin\alpha}{\text{ }gcos\theta} \\ & x - \text{~}\text{direction motion (Taking relative terms }\text{w.r.t.}\text{ box)}\text{~} \\ & u_{x} = + ucos\alpha;a_{x} = 0 \\ & {\text{ }s}_{x} = u_{y}t + \frac{1}{2}a_{y}t^{2} \Rightarrow {\text{ }s}_{x} = ucos\alpha \times \frac{2usin\alpha}{\text{ }gcos\theta} = \frac{u^{2}sin2\alpha}{\text{ }gcos\theta} \end{matrix}\]

(b) For the observer (on ground) to see the horizontal displacement to be zero, the distance travelled by box in time \(\left( \frac{2usin2\alpha}{\text{ }gcos\theta} \right)\) should be equal to the range of the particle w.r.t. box.

Let the speed of the box at the time projection of particle be U . Then for the motion of box with respect to ground.

\[\begin{matrix} & u_{x} = - U;a_{x} = - gsin\theta;t = \frac{2usin\alpha}{\text{ }gcos\theta};s_{x} = \frac{- u^{2}sin2\alpha}{\text{ }gcos\theta} \\ & {\text{ }s}_{x} = u_{x}t + \frac{1}{2}a_{x}t^{2} \\ & \frac{- u^{2}sin2\alpha}{\text{ }gcos\theta} = - U\left( \frac{2usin\alpha}{\text{ }gcos\theta} \right) - \frac{1}{2}\text{ }gsin\theta\left( \frac{2usin\alpha}{\text{ }gcos\theta} \right)^{2} \end{matrix}\]

on solving we get

\[U = \frac{ucos(\alpha + \theta)}{cos\theta}\]

Q. 7 A man wants to cross a river 500 m wide. Rowing speed of the man relative to water is \(3\text{ }km/hr\) and river flows at the speed of \(2\text{ }km/hr\). If man's walking speed on the shore is \(5\text{ }km/hr\), then in which direction he should start rowing in order to reach the directly opposite point on the other bank in shortest time.
Sol. Let he sould start at an angle \(\theta\) with the normal hence

\[{\overrightarrow{v}}_{m} = (u - vsin\theta)\widehat{i} + vcos\widehat{j}\]

Here \({\overrightarrow{v}}_{m} =\) velocity of the man relative to ground.

\[\begin{matrix} & v = \text{~}\text{velocity of the man relative to water}\text{~} \\ & u = \text{~}\text{velocity of water}\text{~} \end{matrix}\]

Hence time taken by the man to cross the river is \(t_{1} = \frac{0.5}{vcos\theta}\)

\(\therefore\ \) Drift of the man along the river is

\[\begin{matrix} & x = (u - vsin\theta)t_{1} \\ & x = (u - vsin\theta)\frac{0.5}{vcos\theta} \end{matrix}\]

Time taken by the man to cover this distance is

Therefore, time of collision is

\[t = \frac{AB}{\left| {\overrightarrow{u}}_{AB} \right|} = \frac{100}{30\sqrt{3} + 40}\]

or

\[t = 1.09\text{ }s\]

(c) Distance of point \(P\) from \(A\) where collision takes place is

\[\begin{matrix} s & \ = \sqrt{\left( u_{Ax}t \right)^{2} + \left( u_{Ay}t - \frac{1}{2}gt^{2} \right)^{2}} \\ & \ = \sqrt{(30\sqrt{3} \times 1.09)^{2} + \left( 30 \times 1.09 - \frac{1}{2} \times 10 \times 1.09 \times 1.09 \right)^{2}} \\ s & \ = 62.64\text{ }m \end{matrix}\]

\[t_{2} = \frac{0.5\left( \frac{usec\theta}{v}tan\theta \right)}{5} = 0.1\left( \frac{u}{v}sec\theta - tan\theta \right)\]

Therefore,
total time \(T = t_{1} + t_{2}\)

\[\Rightarrow \ T = \frac{0.5}{v}sec\theta + \frac{0.1u}{v}sec\theta - 0.1tan\theta\]

Putting the value of \(u\) and \(v\), we get

\[\begin{matrix} & T = \frac{0.5}{3}sec\theta + \frac{0.1 \times 2}{3}sec\theta - 0.1tan\theta \\ & = \frac{0.7}{3}secsec\theta - 0.1tan\theta \\ \Rightarrow \ & \frac{dT}{\text{ }d\theta} = \frac{0.7}{3}sec\theta tan\theta - 0.1\sec^{2}\theta \\ & \text{~}\text{for T to be minimum}\text{~} \\ & \frac{dT}{\text{ }d\theta} = 0 \\ \Rightarrow \ & sin\theta = (3/7) \\ \Rightarrow \ & \theta = \sin^{- 1}(3/7) \end{matrix}\]