Name: Understanding Speed-Time Graph
Uploaded: 2019-07-19
Duration: 2 min 44 s
Description: Any idea, information or concept represented graphically is easier to understand than an equation or other mathematical form. Watch this video to realise this through the speed-time graph.

Question

Derive equations of motion by two different methods.

Accepted Answer

The equations of motion can be derived using the graphical method and by using the formula.

Graphical method:
The velocity $-$ time graph of an object moving with uniform acceleration is as under:

Here,
The initial velocity of the object is given by $u = O A = P M$ .
The final velocity of the object is given by $v = O N = P B$ .
Time taken is given by $t = O P = A M = N B$ .

Derivation of the first equation of motion:
We can find the acceleration ( $a$ ) of the object from the above graph which is given by:
$a =$ slope of the line $A B$ $= \frac{B M}{A M}$
From the graph, we know that, $B M = P B - P M = v - u$ and $A M = t$ .

Substituting these values in the above equation we get,
$a = \frac{v - u}{t} \Rightarrow a t = v - u \Rightarrow v = u + a t$
Thus, $v = u + a t$ is the first equation of motion.

Derivation of second equation of motion:
The area under the velocity $-$ graph gives the displacement ( $s$ ) of the object in time $t$ .
$∴ s =$ area under the graph $A B$ $=$ area of the rectangle $O A M P$ $+$ area of triangle $A B M$ .

Substituting values from the graph we get:
$s = (O A \times O P) + (\frac{1}{2} \times B M \times A M) = u \times t + \frac{1}{2} (v - u) \times t (∵ B M = P B - P M = v - u)$
But from the first equation of motion, we get $v - u = a t$ .
$∴ s = u t + \frac{1}{2} (a t) t = u t + \frac{1}{2} a t^{2}$ .
Thus, $s = u t + \frac{1}{2} a t^{2}$ is the second equation of motion.

Derivation of third equation of motion:
The area under the velocity $-$ graph gives the displacement of the object in time $t$ .
$∴ s =$ area of the trapezium $O A B P$ $= \frac{1}{2} \times (P B + O A) \times O P$ .
Substituting values from the graph we get:
$s = \frac{1}{2} (v + u) \times t$ .
But from the first equation of motion, we have got $t = \frac{(v - u)}{a}$
$∴ s = \frac{1}{2} \times (v + u) \times \frac{(v - u)}{a} \Rightarrow 2 a s = (v + u) \times (v - u) = v^{2} - u^{2}$ .
Thus, $2 a s = v^{2} - u^{2}$ is the third equation of motion.

Deriving the equations of motion using formula:

Derivation of the first equation of motion:
We know that the acceleration is $a = \frac{v - u}{t}$ .
$∴ a t = v - u$
Thus, $v = u + a t$ is the first equation of motion.

Derivation of the second equation of motion:
For uniformly accelerated motion, $v_{a v} = \frac{u + v}{2}$ and $s = v_{a v} \times t$
$∴ s = \frac{u + v}{2} \times t$
Now substituting $v = u + a t$ in the above equation we get,
$s = \frac{u + u + a t}{2} \times t = \frac{2 u + a t}{2} \times t = (\frac{2 u}{2} \times t) + (\frac{a t}{2} \times t) = u t + \frac{1}{2} a t^{2}$ .
Thus, $s = u t + \frac{1}{2} a t^{2}$ is the second equation of motion.

Derivation of the third equation of motion:
From the first equation of motion, we get $t = \frac{v - u}{a}$ .
Substituting the value of $t$ in the second equation of motion we get,
$s = u (\frac{v - u}{a}) + \frac{1}{2} a (\frac{v - u}{a})^{2} = \frac{u v - u^{2}}{a} + \frac{1}{2} \times a \times \frac{v^{2} - 2 v u + u^{2}}{a^{2}}$
$∴ s = \frac{v u - u^{2}}{a} + \frac{v^{2} - 2 v u + u^{2}}{2 a} = \frac{2 v u - 2 u^{2} + v^{2} - 2 v u + u^{2}}{2 a}$
$∴ s = \frac{v^{2} - u^{2}}{2 a} \Rightarrow 2 a s = v^{2} - u^{2}$ .
Thus, $2 a s = v^{2} - u^{2}$ is the third equation of motion.

Hence, all the three equations of motion has been derived by using both the graphical method and by using the formula.

Derive equations of motion by two different methods.

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