Question

The position of a particle varies with time according to the equation $x\left(t\right)=3t^3-4t^2+2$.
The acceleration of the particle at $t=2s$ is:

• A. $44m/s^2$
• B. $28m/s^2$
• C. $20m/s^2$
• D. $16m/s^2$
• E. $36m/s^2$

Answer 1

Answer: (B)

$28m/s^2$

Position of a particle $x\left(t\right)=3t^3-4t^2+2$ m

Differentiating w.r.t time we get $v\left(t\right)=\frac{dx\left(t\right)}{dt}=9t^2-8t$ $m/s$

Thus acceleration of the particle $a\left(t\right)=\frac{dv\left(t\right)}{dt}=18t-8$ $m/s^2$

$⟹$ $a{|}_{t=2s}=18\times 2-8=28$ $m/s^2$