Question

Draw velocity-time graph for a uniformly accelerated object. Derive second and third equations of motion by graphical method.

Accepted Answer

Velocity-Time Graph

A graph that shows the variation of velocity with time is called a velocity-time graph. In this graph, time is taken along x-axis and velocity is taken along the y-axis. The body travels along a straight line and its velocity changes by equal intervals in an equal amount of time for a uniformly accelerated motion.

Below graph clearly mentions the difference between a body initially at rest and a body initially not at rest.

Derivation of second equation of motion using velocity-time graph:

Consider a velocity-time graph for a body that moves with uniform acceleration. The body has an initial velocity $u (u \neq 0)$ at point $A$ and then the body accelerates uniformly in time $t$ to acquire a final velocity $v$ at point $B$ .

The body is moving with uniform acceleration and travels a distance $s$ in time $t$ . The distance travelled by the body is given by the area under the velocity-time graph $A B$ which is equal to the area of trapezium $O A B C$ . Thus,

$D i s \tan c e t r a v e l l e d (s) = A r e a o f t r a p e z i u m O A B C s = A r e a o f t r i a n g l e A B D + A r e a o f r e c \tan g l e O A D C s = (\frac{1}{2} \times b a s e \times h e i g h t) + (l e n g t h \times b r e a d t h) s = (\frac{1}{2} \times A D \times B D) + (O C \times O A) . . . . . . . . . . . . . . . . . (1)$

From the graph,

$A D = O C = t O A = D C = u B D = (B C - D C) B D = v - u B D = u + a t - u [S i n c e v = u + a t] B D = a t$

Substituting the above values in equation (1), we get

$s = (\frac{1}{2} \times t \times a t) + (t \times u) s = \frac{1}{2} a t^{2} + u t s = u t + \frac{1}{2} a t^{2}$

This is the second equation of motion.

Derivation of third equation of motion using velocity-time graph:

Using the same graph, distance travelled, $s$ is given by area of trapezium $O A B C$

$D i s \tan c e t r a v e l l e d = A r e a o f t r a p e z i u m O A B C s = [\frac{1}{2} \times s u m o f p a r a l l e l s i d e s \times h e i g h t] s = \frac{1}{2} \times (O A + B C) \times O C$

Substituting the values in above equation,

$s = \frac{1}{2} (u + v) t$

We know,

$v = u + a t t = \frac{v - u}{t}$

Therefore,

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

This is the third equation of motion.

Draw velocity-time graph for a uniformly accelerated object. Derive second and third equations of motion by graphical method.

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