An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $16 K J$ $16 KJ$ to $96 K J$ $96KJ$ over $t \in [0, 6 s]$ $t in [0, 6 s]$ . What is the average speed of the object?

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Answer 1 · 2017-04-27T23:58:08

163.345

Here, let's assume that uniform change means that the graph of Kinetic energy vs. time is a straight line.

Let y coordinate be kinetic energy, and x coordinate be time. As at x=0, y=16000, so y-intercept of line is 16000.

Slope of the line is $\frac{96000 - 16000}{6 - 0} = \frac{40000}{3}$ $(96000-16000)/(6-0)=40000/3$

Equation of line is
$y = (\frac{40000}{3}) x + 16000$ $y = (40000/3)x + 16000$

$K . E . = (\frac{40000}{3}) t + 16000$ $K.E. = (40000/3)t + 16000$

$\frac{1}{2} m v^{2} = (\frac{40000}{3}) t + 16000$ $1/2 mv^2 = (40000/3)t + 16000$
Putting value of m as 4 and simplifying:

$v^{2} = (\frac{20000}{3}) t + 8000$ $v^2 = (20000/3)t + 8000$
$v = \sqrt{(\frac{20000}{3}) t + 8000}$ $v = sqrt((20000/3)t + 8000)$

Putting $v = \frac{d s}{d t}$ $v=(ds)/dt$ , taking dt to R.H.S and integrating from t=0 to t=6, we get

$s = \int_{0}^{6} (\sqrt{(\frac{20000}{3}) t + 8000}) d t$ $s = int_0^6(sqrt((20000/3)t + 8000))dt$

Upon integrating (which I leave up to you), $s \approx 980.07$ $s~~980.07$

We know that average speed = total distance / total time.

So Avg. Speed $\approx \frac{980.07}{6}$ $~~980.07/6$

Avg. Speed $\approx 163.345 \frac{m}{s}$ $~~163.345 m/s$

An object has a mass of 4 kg4kg4 kg. The object's kinetic energy uniformly changes from 16 KJ16KJ16 KJ to 96KJ96KJ 96KJ over t in [0, 6 s]t∈[0,6s]t in [0, 6 s]. What is the average speed of the object?