Question

Velocity of the particle in a rectilinear motion is given as $v=3t^3-6t$ in $\text{m/s}$. Find
(i) Acceleration at $3$ sec.
(ii) Average acceleration in $3$ sec.

• A. (i) $75m/s^2$
  (ii) $21m/s^2$
• B. (i) $65m/s^2$
  (ii) $21m/s^2$
• C. (i) $75m/s^2$
  (ii) $25m/s^2$
• D. (i) $70m/s^2$
  (ii) $28m/s^2$

Answer 1

The correct option is A (i) $75m/s^2$

(ii) $21m/s^2$
Given, $v=3t^3-6t$
(i) Acceleration at $3$ sec means we have to find out the instantaneous acceleration at $3$ sec
$\Rightarrow a\left(t\right)=\frac{dv\left(t\right)}{dt}=\frac{d(3t^3-6t)}{dt}=9t^2-6$
$\Rightarrow a{|}_{@t=3sec}=9\times 3^2-6=75m/s^2$

(ii) Average acceleration in $3$ sec is given by
$\overline{a}=\frac{(v_f-v_i)}{\Delta t}$
$v_f{|}_{@t=3sec}=3\times 3^3-6\times 3=63\text{m/s}$
$v_i{|}_{@t=0sec}=3\times 0^3-6\times 0=0\text{m/s}$
$\overline{a}=\frac{(v_f-v_i)}{\Delta t}=\frac{63-0}{3}=21m/s^2$