Question

Given below is a velocity-time graph. With reference to the graph, which of the following statement is correct?

• A. The area under the given curve is $\frac{1}{2}\left(v+v_{0}\right) t-v_{0} t$.
• B. The area under the given curve is $\frac{1}{2}\left(v+v_{0}\right) t+v_{0} t .$
• C. The displacement of the object in terms of $v_{0}$ and $a$ is $x=v_{0} t+\frac{1}{2} a t^{2}$.
• D. The relation $x=v_{0} t+\frac{1}{2} a t^{2}$ is same as $x=\left(\frac{v-v_{0}}{2}\right) t$

Answer 1

Answer: (C)

Displacement of the object = Area under $v$ - $t$ curve.
$x=\frac{1}{2}\left(v-v_{0}\right) t+v_{0} t$ ....(i)
Also, $v=v_{0}+a t \Rightarrow v-v_{0}=a t$ ....(ii)
Putting the value of $\left(v-v_{0}\right)$ from Eq. (ii) in Eq. (i), we get
$x =\frac{1}{2} a t \times t+v_{0} t$
$\Rightarrow x =\frac{1}{2} a t^{2}+v_{0} t$
Thus, the statement given in option (c) is correct, rest are incorrect.