Question

Velocity of a particle varies as -
$v=2t^3-3t^2$ in $km/hr$
If $;t=0$ is taken at $12:00$ noon, t is in hr
Find the time between $12:00$ noon and $1:00pm$ at which speed is maximum.

• A. $12:00$ noon
• B. $1:00pm$
• C. $11:00am$
• D. $2:00pm$

Answer 1

The correct option is B $1:00pm$

It is given that $v\left(t\right)=2t^3-3t^2$.

At maximum velocity, the derivative of the velocity with respect to time.

$\begin{array}{c}\frac{dv\left(t\right)}{dt}=0\\ Therefore,\\ \frac{dv\left(t\right)}{dt}=6t^2-6t=0\\ 6t\left(t-1\right)=0\\ t=0ort=1\end{array}$

It is given that $t=0$ is taken at $12:00$ noon.

Therefore ,time at maximum speed $=12:000+1$hour$=13:00$ pm.