Question

A car moves along a straight line, whose equation of motion is given by $s=12t+3t^2-2t^3$ where $s$ is in meters and $t$ in seconds. The velocity of car at the start will be

• A. $7m{s}^{-1}$
• B. $9m{s}^{-1}$
• C. $12m{s}^{-1}$
• D. $16m{s}^{-1}$

Answer 1

The correct option is C $12m{s}^{-1}$

We have given,

distance, $s=12t+3t^2-2t^3$

$\because v=\frac{ds}{dt}$

$\Rightarrow v=\frac{d}{dt}(12t+3t^2-2t^3)$

$\Rightarrow v=12+6t-6t^2$

So, at $t=0$

$v=12m{s}^{-1}$