Question

If $s=2t^3-6t^2+at+5$ is the distance traveled by a particle at time $t$ and if the velocity is $-3$ when its acceleration is zero, then the value of $a$ is

• A. $-3$
• B. $3$
• C. $4$
• D. $-4$
• E. $2$

Answer 1

Answer: (B)

$3$

Given, $s=2t^3-6t^2+at+5$
Velocity, $V=\frac{ds}{dt}=6t^2-12t+a$
Acceleration, $A=\frac{d^{2}s}{d{t}^2}=12t-12$
Since, $A=0$
$\Rightarrow 12t-12=0$
$\Rightarrow t=1$
At $t=1$,
$V=6-12+a=a-6$
It is given that, at $A=0$, $V=-3$
$\Rightarrow a-6=-3$
$\Rightarrow a=3$