# An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 32 KJ to 72 KJ over t in [0, 4 s]. What is the average speed of the object?

Apr 8, 2016

$\text{Initial KE} = \left({E}_{i}\right) = 32 k J = 32 \times {10}^{3} J$

After 4s

$\text{Final KE} = \left({E}_{f}\right) = 72 k J = 72 \times {10}^{3} J$

KE uniformly increases, so it will have a constant positive slope when plotted against time and this slope will be $\frac{\Delta E}{\Delta t} = \frac{{E}_{f} - {E}_{i}}{\Delta t} = \frac{\left(72 - 32\right) \times {10}^{3}}{4} = {10}^{4} \frac{J}{s}$

$, \text{where } \Delta t = 4 s$

If at t th instant ($t \in \left[0 , 4\right]$) the velocity of the mass be $v$ then the KE at $t$ th instant

${E}_{t} = \frac{1}{2} \times m \times {v}^{2} = \frac{1}{2} \times 2 \times {v}^{2} = {v}^{2}$

$\text{,where mass } m = 2 k g$

Now $\frac{{E}_{t} - {E}_{i}}{t - 0} = \frac{{v}^{2} - 32 \times {10}^{3}}{t - 0} = {10}^{4}$

$\implies {v}^{2} = {10}^{4} \times t + 32 \times {10}^{3}$

$\implies v = \sqrt{{10}^{4} \times t + 32 \times {10}^{3}}$

$= \sqrt{{10}^{4} \left(t + 3.2\right)}$

$= {10}^{2} \sqrt{t + 3.2}$

Now distance traversed during 4 s

$S = {\int}_{0}^{4} v \mathrm{dt} = {\int}_{0}^{4} {10}^{2} \left(\sqrt{t + 3.2}\right) \mathrm{dt}$

Let ${z}^{2} = t + 3.2$

$2 z \mathrm{dz} = \mathrm{dt}$

when  t=0; then $z = \sqrt{3.2}$
when  t=4; then $z = \sqrt{7.2}$

$S = {\int}_{0}^{4} v \mathrm{dt} = {\int}_{0}^{4} {10}^{2} \left(\sqrt{t + 3.2}\right) \mathrm{dt}$

$S = {\int}_{\sqrt{3.2}}^{\sqrt{7.2}} {10}^{2} \times 2 {z}^{2} \mathrm{dz}$

$= {\int}_{\sqrt{3.2}}^{\sqrt{7.2}} 200 \times {z}^{2} \mathrm{dz}$

$= \frac{200}{3} \times \left[{\left(\sqrt{7.2}\right)}^{3} - {\left(\sqrt{3.2}\right)}^{3}\right] m$

Hence the average speed $= \text{distance" /"time} = \frac{S}{4}$

$= \frac{200}{3} \times \frac{{\left(\sqrt{7.2}\right)}^{3} - {\left(\sqrt{3.2}\right)}^{3}}{4} = 226.6 \frac{m}{s}$