Question

Derive the equations of motion by Calculus method.

Equations :

v = u + at

s = ut + 1 / 2 at^2

v^2 - u^2 = 2as

Answer 1

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⏩Refer to the following attachment:-

Answer 2

The equations of motion apply to bodies that are moving and these are called as kinematic equations of a body in motion.

To derive the equations we need to use the concept of differentiation and integration.

1st equation:

The quantities of velocity (v), time (t) and distance (s) are related to one another by the equation given as:

$v=\frac{s}{t}$

With a small change in the distance with respect to time we have:

$v=\frac{ds}{dt} \;-------(1)$

When a velocity time graph is drawn then it is said that the slope of the graph corresponds to the acceleration of the body. From the graph, we know that acceleration is defined to be the change in velocity with change in the time.

Acceleration comes into picture only when there is variable velocity and thus a small change in the velocity with respect to its corresponding time quantity will be given by the equation:

$a=\frac{dv}{dt} \;-------(2)$

By rearranging the terms of equation (2) we have:

$adt=dv$

This equation gives a small change in velocity with time but in-order to compute the total velocity of the body with respect to time we need to integrate each small value of velocity.

When we integrate on both sides of the above equation we get:

$a\int\limits^t_0 \, dt =\int\limits^v_u \, dv$

The acceleration (a) is taken out as it is a constant. The time period is integrated over the initial time when the body was at rest taken to be 0 s over the entire journey, that is, over time period t.

Here the limits that have been taken are the initial (u) and the final velocity (v) because the total velocity of the body is given by the difference in the final and the initial velocities of the body.

$\implies a[t]_0^t=[v]_u^v$

$\implies a[t-0]=[v-u]$

$\implies at=v-u$

On rearranging the terms we have:

$v = u + at \;-------(3)$

This is the first equation of kinematics.

2nd equation:

From equation (1) we have:

$v=\frac{ds}{dt}$

$\implies ds= v dt$

On substituting the equation one of motion that has been derived for final velocity we have:

$\implies ds=(u + at) dt$  [∵ From (3),  $v = u + at$]

On solving further we have:

$\implies ds=u\;dt+at\;dt$

By applying integration on both sides in-order to find the total distance we have:

$\int\limits^s_0 \, ds= u\int\limits^t_0 \, dt+ a\int\limits^t_0 {t} \, dt$

The time period is integrated over the same limits or intervals while the distance covered is initially taken to be 0 km and then integrated over total distance s.

By applying the upper and lower limits we get:

$\implies [s]_0^s=[t]_0^t+a[\frac{t^{2} }{2} ]_0^t$     [∵$\int {n} \, dn=\frac{n^{2} }{2}$]

$\implies [s-0]=u[t-0]+a[\frac{t^{2} }{2} -0]$

$\implies s=ut+a[\frac{t}{2} ]$

This is the second equation of kinematics.

3rd equation:

To derive the 3rd equation the equations (1) and (2) are required. Thus we have:

We know that, $v=\frac{ds}{dt}$ and  $a=\frac{dv}{dt}$

We take equation (2) and try and equate it or relate it with (2). Thus, by multiplying the equation by v on either side will give:

$av=v\frac{dv}{dt}$  

Thus, by substituting equation (1) we get:

$a\frac{ds}{dt}=v\frac{dv}{dt}$  [∵$v=\frac{ds}{dt}$]

The common term dt is cancelled out and we have:

$\implies a\;ds=v\;dv$

On integrating both sides of the equations with the same limits we get:

$\implies a\;\int\limits^s_0 \, ds =\int\limits^v_u {v} \, dv$

On further solving we get:

$\implies a[s]_0^s =[\frac{v^{2} }{2} ]_u^v$

$\implies as =[\frac{v^{2}-u^{2} }{2} ]$

$\implies 2as =v^{2}-u^{2}$

This is the third equation of motion.

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