Question

The velocity of a particle moving along the positive $x$-axis is given by $v\left(t\right)=-t^3+6t^2+2t$. What is the magnitude of maximum acceleration and also find the time when it is attained?

• A. $a_{\text{max}}=16{\text{m/s}}^2,t=12\text{s}$
• B. $a_{\text{max}}=14{\text{m/s}}^2,t=2\text{s}$
• C. $a_{\text{max}}=38{\text{m/s}}^2,t=2\text{s}$
• D. $a_{\text{max}}=14{\text{m/s}}^2,t=12\text{s}$

Answer 1

The correct option is B $a_{\text{max}}=14{\text{m/s}}^2,t=2\text{s}$

Given, $v\left(t\right)=-t^3+6t^2+2t$
Now,
Acceleration, $a=\frac{dv}{dt}$
$a=\frac{d}{dt}(-t^3+6t^2+2t)$
$\Rightarrow a=-3t^2+12t+2$
Applying the condition of maximum acceleration,
$\frac{da}{dt}=0$

$\Rightarrow -6t+12=0$
$\therefore t=2\text{s}$
At $t=2\text{s}$ ,either maxima or minima can exist,depending upon the sign of $2^{nd}$ derivative of acceleration w.r.t time.

$\Rightarrow$ for maximum acceleration second derivative of acceleration with respect to time should be negative.
$\therefore \frac{d^{2}a}{d{t}^2}=-6<0$
$\Rightarrow$(-ve value of $\frac{d^{2}a}{d{t}^2}$ represents that maxima exists for acceleration at $t=2\text{s}$)

putting $t=2\text{s}$ in acceleration equation:
Maximum acceleration; $a_{\text{max}}=(-3t^2+12t+2){|}_{t=2\text{s}}$
$\Rightarrow a_{\text{max}}=14{\text{m/s}}^2$