Question

For given velocity-time graph, which of the following options represents correct acceleration-time graph?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C


From velocity-time graph :
Before $t=t_0$,

Slope $\left(m\right)$ of $v-t$ curve is negative $(\text{because}{\theta }_1>{90}^{\circ })$ and constant.

$\Rightarrow m=a=\frac{dv}{dt}<0$

$\Rightarrow a<0$ (negative) and constant.

Similarly,

After $t=t_0$,

Slope $\left(m\right)$ of $v-t$ curve is positive $({\theta }_2<{90}^{\circ })$ and constant.

$\Rightarrow m=a=\frac{dv}{dt}>0$

$\Rightarrow a>0$ (positive) and constant.

Based on these observations, the $a-t$ graph is

Hence, option $\left(c\right)$ is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept - The slope of}v-t\text{graph at an instant}\\ \text{gives acceleration at that instant.}\\ a=\frac{dv}{dt}=\text{slope of}v-t\text{graph}\\ \text{Tip - Break the graph into sections and read - its}\\ \text{behaviour based on knowledge of kinematics}\end{array}$