Question

A particle starts from rest with a time varying acceleration $a=(2t-4)$. Here $t$ is in $s$ and $a$ in $m/s^2$. At what time, particle will have maximum velocity in negative direction?

• A. $3$
• B. $4$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

Acceleration – time graph of the particle is a straight line as shown in the figure. Area enclosed by the $a-t$ graph and the $x$-axis gives the change in velocity .

Since the area enclosed by $a-t$ graph is negative up to $t=2s$, change in velocity is also negative. Therefore, maximum velocity in negative direction will occur at $t=2s$.