Question

Derive equations of motion by two different methods.

Answer 1

Method 1: Derivation of equations of motion by graphical method

1. Let us consider a motion of a body in a straight line with increasing velocity.
2. Let the body is initially at position A where the initial velocity is$u$.
3. And after time$\left(t\right)$ the body reaches position B where the final velocity is$v$
4. A graph is plotted between velocity and time In which velocity is taken on$y-axis$ and time is taken on$y-axis$.

Step 1:Derivation of the first equation of motion

1. We know that the slope of line AB of the velocity-time graph shows the acceleration of the body in motion.
2. So the acceleration= slope of the line AB $\Rightarrow a=\frac{BM}{AM}$

$\Rightarrow a=\frac{BP-MP}{AM}$$\Rightarrow a=\frac{v-u}{t}$

$\Rightarrow v-u=at$

$\Rightarrow v=u+at$

Step 2:Derivation of the second equation of motion

1. We know that the area under the velocity-time graph shows the distance covered by the body.
2. So the distance = area of the trapezium $OABP$

$\Rightarrow dis\tan ce=s=\frac{1}{2}\times \left(OA+PB\right)\times OP$

$\Rightarrow s=\frac{1}{2}\times \left(u+v\right)\times t$

putting the value of$v$ from the first equation of motion we get

$\Rightarrow s=\frac{1}{2}\left(u+u+at\right)\times t$

$\Rightarrow s=\frac{1}{2}\times 2u\times t+\frac{1}{2}\times at\times t$

$\Rightarrow s=ut+\frac{1}{2}a{t}^2$

Step 3: Derivation of the third equation of motion

As distance is $\Rightarrow s=\frac{1}{2}\times \left(u+v\right)\times t$

from the first equation of motion $v=u+at$$\Rightarrow t=\frac{v-u}{a}$

So $s=\frac{1}{2}\times \left(u+v\right)\left(\frac{v-u}{a}\right)$

$\Rightarrow s=\frac{\left(v+u\right)\left(v-u\right)}{2a}$

$\Rightarrow s=\frac{v^2-u^2}{2a}$

$\Rightarrow v^2-u^2=2as$

Method 2: 2nd method for derivation of equations of motion

Step 1: Derivation of the first equation of motion

1. As per the definition of acceleration, acceleration is the rate of change of velocity with time.
2. Mathematically it can be written as $acceleration=\frac{changeinvelocity}{time}$

If$u=initialvelocity,v=finalvelocity,t=time$

then acceleration$\left(a\right)$ will be$a=\frac{v-u}{t}$

$\Rightarrow v-u=at$

$\Rightarrow v=u+at$

Step 2: Derivation of the second equation of motion

1. If$u=initialvelocity,v=finalvelocity,t=time$, For a uniformly accelerated motion, the average velocity of a body will be$V_{avg}=\frac{u+v}{2}$.
2. We know that relation between displacement$\left(s\right)$, average velocity and time is

$displacement=averagevelocity\times time$

$\Rightarrow s=\frac{u+v}{2}\times t$

from the first equation of motion, the value of $v$ is$v=u+at$

so$s=\left(\frac{u+u+at}{2}\right)\times t$

$\Rightarrow s=\frac{1}{2}\times 2u\times t+\frac{1}{2}\times at\times t$$\Rightarrow s=ut+\frac{1}{2}a{t}^2$

Step 3: Derivation of the third equation of motion

We know that$displacement=averagevelocity\times time$

$\Rightarrow s=\frac{u+v}{2}\times t$

putting the value of t in this equation and simplifying it we can derive the third equation of motion as

$s=\frac{1}{2}\times \left(u+v\right)\left(\frac{v-u}{a}\right)$$\Rightarrow s=\frac{\left(v+u\right)\left(v-u\right)}{2a}$$\Rightarrow s=\frac{v^2-u^2}{2a}$

$\Rightarrow v^2-u^2=2as$

Thus, the equations of motion can be derived by two methods.