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, $v=v_0+at\Rightarrow v-v_0=at$ ....(ii)
Putting the value of $\left(v-v_0\right)$ from Eq. (ii) in Eq. (i), we get
$x=\frac{1}{2}at\times t+v_{0}t$
$\Rightarrow x=\frac{1}{2}a{t}^2+v_{0}t$
Thus, the statement given in option (c) is correct, rest are incorrect.