Please Wait... While Loading Full Video### KINEMATIC EQUATIONS AND RELATIVE VELOCITY

• Kinematic Equations for Uniformly Accelerated Motion

• Relative Velocity

• Relative Velocity

For uniformly accelerated motion, we can derive some simple equations that relate displacement (`x`), time taken (`t`), initial velocity (`v_0`), final velocity (`v`) and acceleration (`a`).

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "First Equation")} `

If the velocity of an object is `v_o` at `t = 0` and `v` at time `t`, we have

`color(blue)(bara=(v-v_o)/(t-0))`

`color(blue)(v=v_o +at)`...................................`(i)`

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Second Equation")} `

The area under this curve is :

Area between instants 0 and t = Area of `"triangle" ABC +` Area of `"rectangle" OACD`

`=1/2(v-v_o)t + v_o t`

`color(green)("We know, the area under v-t curve represents the displacement.")` Therefore,

`color(green)(x=1/2(v-v_o)t + v_o t)`

But `v-v_0 = a t` Therefore, `x=1/2 at^2 +v_ot`

`color(blue)(x=v_ot+1/2at^2)`...........................................`(ii)`

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Third Equation")} `

As we discussed earlier, `x=1/2(v-v_o)t + v_o t`

This equation can also be written as

`x=(v+v_0)/2 t=barv t`.......(a)

where, `barv=(v+v_o)/2`

From first kinematic equation `t=(v-v_o)/a`

Substituting this in Eq. (a), we get

`x=barv t=((v+v_o)/2)((v-v_o)/a)=(v^2-v_o^2)/2`

`color(blue)(v^2=v_o^2 +2ax)`..........................................`(iii)`

Thus, `ul"we have obtained three important equations :"`

`color(blue)(v=v_o +at)`

`color(blue)(x=v_ot+1/2at^2)`

`color(blue)(v^2=v_o^2 +2ax)`

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "First Equation")} `

If the velocity of an object is `v_o` at `t = 0` and `v` at time `t`, we have

`color(blue)(bara=(v-v_o)/(t-0))`

`color(blue)(v=v_o +at)`...................................`(i)`

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Second Equation")} `

The area under this curve is :

Area between instants 0 and t = Area of `"triangle" ABC +` Area of `"rectangle" OACD`

`=1/2(v-v_o)t + v_o t`

`color(green)("We know, the area under v-t curve represents the displacement.")` Therefore,

`color(green)(x=1/2(v-v_o)t + v_o t)`

But `v-v_0 = a t` Therefore, `x=1/2 at^2 +v_ot`

`color(blue)(x=v_ot+1/2at^2)`...........................................`(ii)`

`\color{fuchsia} ★ \color{fuchsia} ul{\mathbf( "Third Equation")} `

As we discussed earlier, `x=1/2(v-v_o)t + v_o t`

This equation can also be written as

`x=(v+v_0)/2 t=barv t`.......(a)

where, `barv=(v+v_o)/2`

From first kinematic equation `t=(v-v_o)/a`

Substituting this in Eq. (a), we get

`x=barv t=((v+v_o)/2)((v-v_o)/a)=(v^2-v_o^2)/2`

`color(blue)(v^2=v_o^2 +2ax)`..........................................`(iii)`

Thus, `ul"we have obtained three important equations :"`

`color(blue)(v=v_o +at)`

`color(blue)(x=v_ot+1/2at^2)`

`color(blue)(v^2=v_o^2 +2ax)`

Q 3043556443

Obtain equations of motion for constant acceleration using method of calculus.

By definition `a = (dv)/(dt)`

`dv = a dt`

Integrating both sides

`int_(v_0)^(v) dv = int_(0)^(t) a dt`

`= a int_(0)^(t) dt` (a is constant)

`v - v_0 = at`

`v = v_0 + a t`

Further, `dx = v dt`

Integrating both sides `int_(x_0)^(x) dx = int_(0)^(t) vdt`

`= int_(0)^(t) (v_0+at) dt`

`x - x_0 = v_0 t +1/2 a t^2`

`x = x_0 +v_0 t +1/2 a t^2`

We can write `a = (dv)/(dt) = (dv)/(dx) (dx)/(dt) = v (dv)/(dx)`

or `vdv = a dx`

Integrating both sides,

`int_(v_0)^(v) vdv = int_(x_0)^(x) adx`

`(v^2-v_0^2)/2 = a(x-x_0)`

`v^2 = v_0^2+2a (x - x_0)`

Q 3033756642

A ball is thrown vertically upwards with a velocity of `20 m s^(–1)` from the top of a multistorey building. The height of the point from where the ball is thrown is `25.0 m` from the ground. (a) How high will the ball rise ? and (b) how long will it be before the ball hits the ground? Take `g = 10 m s^(–2)`.

(a) Let us take the y-axis in the vertically upward direction with zero at the ground, as shown in Fig.

Now `v_0 = +20 ms^(-1)`

` a = -g = -10 ms^(-2)`

`v = 0 ms^(-1)`

If the ball rises to height y from the point of launch, then using the equation `v^2 = v_0^2+2a(y-y_0)`

we get `0 = (20)^2+2(-10) (y-y_0)`

Solving, we get, `(y – y0) = 20 m.`

(b) We can solve this part of the problem in two ways. Note carefully the methods used.

FIRST METHOD : In the first method, we split the path in two parts : the upward motion (A to B) and the downward motion (B to C) and calculate the corresponding time taken `t_1` and `t_2`. Since the velocity at B is zero, we have :

`v = v_0+at`

`0 = 20-10t_1`

or `t_1 = 2s`

This is the time in going from A to B. From B, or the point of the maximum height, the ball falls freely under the acceleration due to gravity. The ball is moving in negative y direction. We use equation.

`y = y_0 +v_0 t +1/2 a t^2`

We have, `y_0 = 45 m , y = 0 , v_0 = 0 , a = -g = -10 ms^(-2)`

`0 = 45 +(1/2) (-10) t_2^2`

Solving, we get t2 = 3 s

Therefore, the total time taken by the ball before it hits the ground `= t_1 + t_2 = 2 s + 3 s = 5 s.`

SECOND METHOD : The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation `y = y_0+v_0 t +1/2 a t^2`

Now `y_0 = 25 m , y = 0 m`

`v_0 = 20 m s^(-1) , a = -10 ms^(-2) , t= ?`

`0 = 25 +20 t + (½) (-10) t^2`

Or, `5t^2 – 20t – 25 = 0`

Solving this quadratic equation for t, we get

`t = 5s`

Q 3023856741

Free-fall : Discuss the motion of an object under free fall. Neglect air resistance.

An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is said to be in free fall. If the height through which the object falls is small compared to the earth’s radius, g can be taken to be constant, equal to `9.8 m s^(–2)`. Free fall is thus a case of motion with uniform acceleration. We assume that the motion is in y-direction, more correctly in –y-direction because we choose upward direction as positive. Since the acceleration due to gravity is always downward, it is in the negative direction and we have

`a = -9 = -9.8 ms^(-2)`

The object is released from rest at `y = 0.` Therefore, `v_0 = 0` and the equations of motion become:

`v = 0- g t \ \ \ \ \ \ = -9.8 t ms^(-1)`

`y = 0-1/2 g t^2 \ \ \ \ \ \ = -4.9 t^2 m`

`v^2 = 0-2 g y \ \ \ \ \ = -19.6 y m^2 s^(-2)`

These equations give the velocity and the distance travelled as a function of time and also the variation of velocity with distance. The variation of acceleration, velocity, and distance, with time have been plotted in Fig. (a), (b) and (c).

Q 3003856748

Galileo’s law of odd numbers : “The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7…...].” Prove it.

Let us divide the time interval of motion of an object under free fall into many equal intervals τ and find out the distances traversed during successive intervals of time. Since initial velocity is zero, we have `y = -1/2 g t^2`

Using this equation, we can calculate the position of the object after different time intervals, `0, τ, 2τ, 3τ`… which are given in second column of Table If we take `(–1/ 2) gτ^2` as `y_0 —` the position coordinate after first time interval `τ`, then third column gives the positions in the unit of `y_o`. The fourth column gives the distances traversed in successive `τs`. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11… as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall.

Q 3043056843

Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity `(v_0)` and the braking capacity, or deceleration, `–a` that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of `v_o` and a.

Let the distance travelled by the vehicle before it stops be `ds`. Then, using equation of motion `v^2 = v_o 2 + 2 ax`, and noting that `v = 0`, we have the stopping distance Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration). For the car of a particular make, the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15, 20 and 25 m/s which are nearly consistent with the above formula.

Q 3083056847

Reaction time : When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on complexity of the situation and on an individual

You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger (Fig. ). After you catch it, find the distance d travelled by the ruler. In a particular case, d was found to be 21.0 cm. Estimate reaction time.

You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger (Fig. ). After you catch it, find the distance d travelled by the ruler. In a particular case, d was found to be 21.0 cm. Estimate reaction time.

The ruler drops under free fall. Therefore, `v_o = 0,` and `g = –9.8 m s^(–2).` The distance travelled `d` and the reaction time `t_r` are related by

`d = -1/2 g t_r^2`

or `t_r = sqrt((2d)/g s)`

Given `d = 21.0 cm` and `g = 9.8 m s^(–2)` the reaction time is `t_r = sqrt((2xx0.21)/(9.8)) s approx 0.2 s`

Consider two objects `A` and `B` moving uniformly with average velocities `v_A` and `v_B` in one dimension, say along x-axis.

If `x_A(0)` and `x_B(0)` are positions of objects `A` and `B`, respectively at time `t = 0,` their positions `x_A (t)` and `x_B(t)` at time `t` are given by:

`x_A (t ) = x_A (0) + v_A t`..................................`(i)`

`x_B (t) = x_B (0) + v_B t`.........................................`(ii)`

Then, the displacement from object `A` to object `B` is given by

`x_(BA)(t) = x_B (t) – x_A (t)`

`x_(BA)(t) = [ x_B (0) – x_A (0) ] + (v_B – v_A) t`..........................................`(iii)`

It tells us that as seen from object `A`, object `B` has a velocity `v_B – v_A` because the displacement from `A` to `B` changes steadily by the amount `v_B – v_A` in each unit of time.

We say that the velocity of object `B` relative to object `A` is `v_B – v_A`

`color(blue)(v_(BA)=v_B-v_A)`

Similarly, velocity of object A relative to object B is:

`color(blue)(v_(AB)=v_A-v_B)`

This shows: `color(blue)(v_(BA) = – v_(AB))`

`color(green)"Now we consider some special cases :"`

`(a)` If `color(blue)(v_B = v_A)`, `\ \ \ \ color(green)(v_B – v_A = 0)`. Then, from Eq. (iii), `x_B (t) – x_A (t) = x_B (0) – x_A (0)`.

Therefore, the two objects stay at a constant distance `(x_B (0) – x_A (0))` apart, and their position–time graphs are straight lines parallel to each other as shown in Fig. The relative velocity `v_(AB)` or `v_(BA)` is zero in this case.

`(b)` If `color(blue)(v_A > v_B)`, `\ \ \ \ color(green)(v_B – v_A) " is negative"`. One graph is steeper than the other and they meet at a common point.

`(c)` In case of `v_A` and `v_B` are of opposite signs.

If `x_A(0)` and `x_B(0)` are positions of objects `A` and `B`, respectively at time `t = 0,` their positions `x_A (t)` and `x_B(t)` at time `t` are given by:

`x_A (t ) = x_A (0) + v_A t`..................................`(i)`

`x_B (t) = x_B (0) + v_B t`.........................................`(ii)`

Then, the displacement from object `A` to object `B` is given by

`x_(BA)(t) = x_B (t) – x_A (t)`

`x_(BA)(t) = [ x_B (0) – x_A (0) ] + (v_B – v_A) t`..........................................`(iii)`

It tells us that as seen from object `A`, object `B` has a velocity `v_B – v_A` because the displacement from `A` to `B` changes steadily by the amount `v_B – v_A` in each unit of time.

We say that the velocity of object `B` relative to object `A` is `v_B – v_A`

`color(blue)(v_(BA)=v_B-v_A)`

Similarly, velocity of object A relative to object B is:

`color(blue)(v_(AB)=v_A-v_B)`

This shows: `color(blue)(v_(BA) = – v_(AB))`

`color(green)"Now we consider some special cases :"`

`(a)` If `color(blue)(v_B = v_A)`, `\ \ \ \ color(green)(v_B – v_A = 0)`. Then, from Eq. (iii), `x_B (t) – x_A (t) = x_B (0) – x_A (0)`.

Therefore, the two objects stay at a constant distance `(x_B (0) – x_A (0))` apart, and their position–time graphs are straight lines parallel to each other as shown in Fig. The relative velocity `v_(AB)` or `v_(BA)` is zero in this case.

`(b)` If `color(blue)(v_A > v_B)`, `\ \ \ \ color(green)(v_B – v_A) " is negative"`. One graph is steeper than the other and they meet at a common point.

`(c)` In case of `v_A` and `v_B` are of opposite signs.

Q 3063156945

Two parallel rail tracks run north-south. Train A moves north with a speed of `54 km h^(–1)`, and train B moves south with a speed of `90 km h^(–1)`. What is the

(a) velocity of B with respect to A ?,

(b) velocity of ground with respect to B ?, and

(c) velocity of a monkey running on the roof of the train A against its motion (with a velocity of `18 km h^(–1)` with respect to the train A) as observed by a man standing on the ground ?

(a) velocity of B with respect to A ?,

(b) velocity of ground with respect to B ?, and

(c) velocity of a monkey running on the roof of the train A against its motion (with a velocity of `18 km h^(–1)` with respect to the train A) as observed by a man standing on the ground ?

Choose the positive direction of x-axis to be from south to north. Then, `v_A = +54 km h^(-1) = 15 m s^(-1)`

`v_B = - 90 km h^(-1) = -25 m s^(-1)`

Relative velocity of `B` with respect to `A = v_B – v_A= – 40 m s^(–1)` , i.e. the train `B` appears to `A` to move with a speed of 40 m s–1 from north to south.

Relative velocity of ground with respect to `B = 0 - v_B = 25 ms^(-1)`

In (c), let the velocity of the monkey with respect to ground be `v_M`. Relative velocity of the monkey with respect to A,

`v_(MA) = v_M – v_A = –18 km h^(–1) =–5 ms^(–1)`. Therefore,

`v_M = (15 – 5) m s^(–1) = 10 m s^(–1).`