# Question #e2a90

Feb 3, 2017

$8 \left(1 + t\right) {\text{ Js}}^{-} 1$

#### Explanation:

We know that Kinetic Energy of an object having mass $m$ and velocity $v$ is given by the expression

$K E = \frac{1}{2} m {v}^{2}$ .....(1)

Also kinematic equation connecting quantities of interest is

$v = u + a t$.....(2)
where $u$ is initial velocity, $a \mathmr{and} t$ are acceleration and time respectively.
Inserting value of $v$ from (2) in (1) we get
$K E = \frac{1}{2} m {\left(u + a t\right)}^{2}$

To find rate of change of kinetic energy we differentiate both sides with respect to time. We get

$\dot{K E} = \frac{d}{\mathrm{dt}} \left[\frac{1}{2} m {\left(u + a t\right)}^{2}\right]$

using chain rule we get
$\dot{K E} = \frac{1}{2} m \left[2 \left(u + a t\right) \times a\right]$

Inserting given values we get
$\dot{K E} = \frac{1}{2} \times 2 \left[2 \left(2 + 2 t\right) \times 2\right]$
$\dot{K E} = 8 \left(1 + t\right) {\text{ Js}}^{-} 1$