Question #e2a90

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Question #e2a90

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Answer 1 · 2017-02-03T04:57:18

$8 (1 + t) {Js}^{- 1}$ $8(1+t)" Js"^-1$

Explanation:

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

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

Also kinematic equation connecting quantities of interest is

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

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

$. K E = \frac{d}{d t} [\frac{1}{2} m {(u + a t)}^{2}]$ $dot(KE)=d/dt[1/2m(u+at)^2]$

using chain rule we get
$. K E = \frac{1}{2} m [2 (u + a t) \times a]$ $dot(KE)=1/2m[2(u+at)xxa]$

Inserting given values we get
$. K E = \frac{1}{2} \times 2 [2 (2 + 2 t) \times 2]$ $dot(KE)=1/2xx2[2(2+2t)xx2]$
$. K E = 8 (1 + t) {Js}^{- 1}$ $dot(KE)=8(1+t)" Js"^-1$

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Question #e2a90

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