An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $96 K J$ $96 KJ$ to $280 K J$ $280 KJ$ over $t \in [0, 12 s]$ $t in [0, 12 s]$ . What is the average speed of the object?

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Answer 1 · 2017-02-16T12:40:25

The average speed is $= 296.65 m s^{- 1}$ $=296.65ms^-1$

The initial kinetic energy is $= \frac{1}{2} m u^{2}$ $=1/2m u^2$

So,

$\frac{1}{2} m u^{2} = 96000$ $1/2m u^2=96000$

$u^{2} = 2 \cdot \frac{96000}{4} = 48000$ $u^2=2*96000/4=48000$

$v = \sqrt{48000} = 219.1 m s^{- 1}$ $v=sqrt48000=219.1ms^-1$

The final kinetic energy is $\frac{1}{2} m v^{2}$ $1/2m v^2$

So,

$\frac{1}{2} m v^{2} = 280000$ $1/2m v^2=280000$

$v^{2} = 2 \cdot \frac{280000}{4} = 140000$ $v^2=2*280000/4=140000$

$v = \sqrt{140000} = 374.2 m s^{- 1}$ $v=sqrt140000=374.2ms^-1$

On a graph velocity v/s time, we have a straight line $D C$ $DC$

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$A D = u$ $AD=u$

$B C = v$ $BC=v$

$A B = t$ $AB=t$

Area of $A B C D = \frac{A D + B C}{2} \cdot t$ $ABCD=(AD+BC)/2*t$

The average speed $= ¯ v$ $=barv$

$¯ v \cdot t = \frac{A D + B C}{2} \cdot t$ $barv*t=(AD+BC)/2*t$

so,

$¯ v = \frac{u + v}{2}$ $barv=(u+v)/2$

$= \frac{219.1 + 374.2}{2}$ $=(219.1+374.2)/2$

$= 296.65 m s^{- 1}$ $=296.65ms^-1$

An object has a mass of 4kg4 kg. The object's kinetic energy uniformly changes from 96KJ96 KJ to 280KJ 280 KJ over t∈[0,12s]t in [0, 12 s]. What is the average speed of the object?