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 expression for the acceleration of the particle.

• A. $3t^2+3t$
• B. $6t(t-1)$
• C. $6t^2+3t$
• D. none of these

Answer 1

The correct option is B $6t(t-1)$

$$\begin{array}{c}Accorrdingtoquestion.......................\\ velocity\left(v\right)=2t^3-3t^2\\ Acceleration=\frac{dv}{dt}\\ \Rightarrow a=\frac{d}{dt}(2t^3-3t^2)\\ \Rightarrow a=2\times 3t^2-3\times 2t=6t^2-6t\\ \therefore a=6t(t-1)\\ SothecorrectoptionisB.\end{array}$$