Question

A graph of acceleration versus time of a particle starting from rest at $t=0$ is as shown in figure. The speed of the particle at $t=14\text{s}$ is

• A. $2{\text{ms}}^{-1}$
• B. $34{\text{ms}}^{-1}$
• C. $20{\text{ms}}^{-1}$
• D. $42{\text{ms}}^{-1}$

Answer 1

The correct option is B $34{\text{ms}}^{-1}$

We know that area under the $a-t$ graph gives change in velocity

Total area under the graph from $t=0$ to $t=14\text{s}$ is given by
$\Rightarrow A=$ Area under the graph from $t=0$ to $t=12\text{s}$ [Trapezium] - Area under the graph from $t=12\text{s}$ to $t=14\text{s}$ [triangle]
$\Rightarrow A=\frac{1}{2}\times 4\times (12+6)-\frac{1}{2}\times 2\times 2$
$\Rightarrow A=36-2=34\text{m/s}$
Now, $A=$ Change in velocity
$A=v_{14}-0$
$v_{14}=34\text{m/s}$