Question

The velocity – time graph of a particle moving along a straight line is as shown in fig. The rate of acceleration and deceleration is constant and it is equal to $5{\text{ms}}^{-2}$. If the average velocity during the motion is $20{\text{ms}}^{-1}$, then

The value of $t$ is

• A. $5\text{s}$
• B. $10\text{s}$
• C. $20\text{s}$
• D. $5\sqrt{2}\text{s}$

Answer 1

The correct option is A $5\text{s}$

Slope of $v-t$ graph = acceleration
From $t=0$ to $t=t$ sec
Slope of line segment = 5
$\Rightarrow \frac{v}{t}=5\Rightarrow v=5t$
Since the velocity is always positive, we can say Displacement = Distance covered
$\Rightarrow$Average velocity = Average speed
$\therefore$ Average speed = $20{\text{ms}}^{-1}$


$\Rightarrow \frac{\text{Total distance}}{\text{Total time}}=20$
$\Rightarrow \frac{\text{Area of graph}}{20+t}=20$
Area under the graph $=\frac{1}{2}\times 5t\times t+5t\times (20-t)+\frac{1}{2}\times 5t\times t=100t$
$\Rightarrow 100t=20(20+t)$
$\Rightarrow t=5\text{s}$