Question

The acceleration-time graph for a particle moving in a straight line is as shown in the figure. Change in the velocity of the particle from $t=0$ to $t=6s$ is

• A. $10m/s$
• B. $4m/s$
• C. $12m/s$
• D. $8m/s$

Answer 1

The correct option is B $4m/s$

Change in velocity = net area under $a-t$ graph
Net area under graph = area of $\Delta OAB$ - area of $\Delta BCD$
$=[\frac{1}{2}\times 4\times 4]-[\frac{1}{2}\times 2\times 4]$
$=8-4$
$=4m/s$