# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 144 KJ to  124KJ over t in [0, 6 s]. What is the average speed of the object?

Mar 26, 2016

I found: $255.8 \approx 256 \frac{m}{s}$

#### Explanation:

From your Kinetic Energies you can work out the speed (at two instants of time) knowing that kinetic energy $K$ is:
$K = \frac{1}{2} m {v}^{2}$
so, rearranging:
${v}_{1} = \sqrt{2 {K}_{1} / m} = \sqrt{\frac{2 \cdot 144 \times {10}^{3}}{4}} = 268 \frac{m}{s}$
${v}_{2} = \sqrt{2 {K}_{2} / m} = \sqrt{\frac{2 \cdot 124 \times {10}^{3}}{4}} = 249 \frac{m}{s}$
Now we use the kinematic relationship:
${v}_{f} = {v}_{i} + a t$

to get:
$249 = 268 + 6 a$
and: $a = - 3.2 \frac{m}{s} ^ 2$
and the other relationship:
${v}_{f}^{2} = {v}_{i}^{2} + 2 a s$

${\left(249\right)}^{2} = {\left(268\right)}^{2} - 2 \cdot \left(3.2\right) s$
$1535 m$

The average speed during your time interval will then be:
${v}_{\text{av}} = \frac{\Delta s}{\Delta t} = \frac{1535}{6} = 255.8 \frac{m}{s}$

ALTERNATIVE
${v}_{f} - {v}_{i} = a t \ldots \ldots \left(1\right)$
${v}_{f}^{2} - {v}_{i}^{2} = 2 a s \ldots \ldots \left(2\right)$

If Eq(2) is divided by Eq (1) we get
$\frac{2 s}{t} = {v}_{f} + {v}_{i}$
av velocity
$: . \frac{s}{t} = \frac{{v}_{f} + {v}_{i}}{2} = 258.5 \frac{m}{s}$

Mar 26, 2016

If Initial velocity be u m/s and final velocity v m/s after 6 s
then$\frac{1}{2} \cdot 4 {u}^{2} = 144000 \implies {u}^{2} = 72000$
Again
$\frac{1}{2} \cdot 4 {v}^{2} = 124000 \implies {v}^{2} = 62000$
average speed for uniform change in velocity( i.e. having uniform retardation) = $\frac{u + v}{2} = 258.6 \frac{m}{s}$

If uniform change in KE with time is considered then calculation will not match with this one

Mar 27, 2016

From the graph $258.7 m {s}^{-} 1$ rounded to one decimal place.

#### Explanation:

It is given that the Kinetic energy changes uniformly.
We know that $K E = \frac{1}{2} m {v}^{2}$.
Divide the given change in velocity by $6$. The time interval of interest.

Calculate velocities at time $t = 0 , + 1 , + 2 , + 3 , + 4 , + 5 , + 6$

Observe that the change in velocity in each second interval is not constant. As such, acceleration is not constant. Therefore, the known expressions for linear motion for constant acceleration can not be applied.

I have attempted to assess Approx Average velocity by calculating distance moved in each second. This is approximately equal to the velocity for that second. Take the sum of these six distances and divide by time that is $6 s$.

More accurately, one should draw a $v \text{ vs } t$ graph. Find out area under this graph to calculate total distance moved.  Mar 28, 2016

~~258.8m/s

#### Explanation:

Initial KE $\left({E}_{i}\right) = 144 k J = 144 \times {10}^{3} J$
After 6s Final KE $\left({E}_{f}\right) = 124 k J = 124 \times {10}^{3} J$
KE is uniformly decreased, so it will have a constant negative slope
and this slope will be $\frac{\Delta E}{\Delta t} = \frac{{E}_{f} - {E}_{i}}{\Delta t} = \frac{\left(124 - 144\right) \times {10}^{3}}{6} = - \frac{10}{3} \times {10}^{3}$ where $\Delta t = 6 s$

If at t th instant $0 < t < 6$ the velocity of the mass be v
We are to find out a relation between v and t
the KE at t th instant ${E}_{t} = \frac{1}{2} \times 4 \times {v}^{2} = 2 {v}^{2}$
Now $\frac{{E}_{t} - {E}_{i}}{t - 0} = \frac{2 {v}^{2} - 144 \times {10}^{3}}{t - 0} = - \frac{10}{3} \times {10}^{3}$
$\implies 2 {v}^{2} = - \frac{10}{3} \times {10}^{3} \times t + 144 \times {10}^{3}$
$\implies {v}^{2} = 72 \times {10}^{3} - \frac{5}{3} \times {10}^{3} \times t$
$\implies v = \sqrt{72 \times {10}^{3} - \frac{5}{3} \times {10}^{3} \times t}$

Now distance traversed during 6 s

$S = {\int}_{0}^{6} v \mathrm{dt} = {\int}_{0}^{6} \left(\sqrt{72 \times {10}^{3} - \frac{5}{3} \times {10}^{3} \times t}\right) \mathrm{dt}$
Let ${z}^{2} = 72 \times {10}^{3} - \frac{5}{3} \times {10}^{3} \times t$
$2 z \mathrm{dz} = - \frac{5}{3} \times {10}^{3} \mathrm{dt}$
$\implies \frac{6}{5} \times {10}^{-} 3 z \mathrm{dz} = - \mathrm{dt}$
when t=0; then z =10sqrt720 
when t=6; then z =10sqrt620 

$S = {\int}_{0}^{6} v \mathrm{dt} = {\int}_{0}^{6} \left(\sqrt{72 \times {10}^{3} - \frac{5}{3} \times {10}^{3} \times t}\right) \mathrm{dt}$

$S = - {\int}_{10 \sqrt{720}}^{10 \sqrt{620}} \frac{6}{5} \times {10}^{-} 3 {z}^{2} \mathrm{dz}$

$= {\int}_{10 \sqrt{620}}^{10 \sqrt{720}} \frac{6}{5} \times {10}^{-} 3 {z}^{2} \mathrm{dz}$

$= \frac{6}{5} \times {10}^{-} 3 \times \frac{1}{3} \times \left[{\left(10 \sqrt{720}\right)}^{3} - {\left(10 \sqrt{620}\right)}^{3}\right]$

$= \frac{6}{5} \times {10}^{-} 3 \times \frac{1}{3} \times \left[{\left({10}^{2} \sqrt{7.2}\right)}^{3} - {\left({10}^{2} \sqrt{6.2}\right)}^{3}\right]$
$= \frac{6}{5} \times {10}^{-} 3 \times \frac{1}{3} \times {10}^{6} \left[{\left(\sqrt{7.2}\right)}^{3} - {\left(1 \sqrt{6.2}\right)}^{3}\right]$
$= \frac{2}{5} \times {10}^{3} \times \left[{\left(\sqrt{7.2}\right)}^{3} - {\left(\sqrt{6.2}\right)}^{3}\right]$
$= \frac{2}{5} \times {10}^{3} \times 3.88175 = 1552.7 m$

Hence the average speed S/6=1552.7/6`~~258.8m/s# 