An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 144 KJ to  120KJ over t in [0, 3 s]. What is the average speed of the object?

Dec 13, 2017

Average speed = $\frac{{v}_{0} + v}{2} \approx 256.64 \frac{m}{s}$

Average acceleration = $\frac{{v}_{0} - v}{3} \approx - 7.794 \frac{m}{{s}^{2}}$

Explanation:

For this question we must cosnider the formula for $K . E$

$K . E = \frac{1}{2} \cdot m \cdot {v}^{2}$

Initial Kinetic energy$\implies$ $K . {E}_{0} = 144 \cdot 1000 J$

$\implies \frac{1}{2} \cdot 4 \cdot {v}_{0}^{2} = 144000$

$\implies {v}_{0} \approx 268.33 \frac{m}{s}$

Final Kinetic energy, $\implies K . E = 120 \cdot 1000 J$

$\implies \frac{1}{2} \cdot 4 \cdot {v}^{2} = 120000$

$\implies v \approx 244.95 \frac{m}{s}$

Average speed = $\frac{{v}_{0} + v}{2} \approx 256.64 \frac{m}{s}$

Average acceleration = $\frac{{v}_{0} - v}{3} \approx - 7.794 \frac{m}{{s}^{2}}$

Dec 13, 2017

The average speed is $= 256.8 m {s}^{-} 1$

Explanation:

The kinetic energy is

$K E = \frac{1}{2} m {v}^{2}$

The mass is $m = 4 k g$

The initial velocity is $= {u}_{1} m {s}^{-} 1$

The final velocity is $= {u}_{2} m {s}^{-} 1$

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 144000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 120000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 144000 = 72000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 120000 = 60000 {m}^{2} {s}^{-} 2$

The graph of ${v}^{2} = f \left(t\right)$ is a straight line

The points are $\left(0 , 72000\right)$ and $\left(3 , 60000\right)$

The equation of the line is

${v}^{2} - 72000 = \frac{60000 - 72000}{3} t$

${v}^{2} = - 4000 t + 72000$

So,

$v = \sqrt{- 4000 t + 72000}$

We need to calculate the average value of $v$ over $t \in \left[0 , 3\right]$

$\left(3 - 0\right) \overline{v} = {\int}_{0}^{3} \left(\sqrt{- 4000 t + 72000}\right) \mathrm{dt}$

$3 \overline{v} = {\left(- 4000 t + 72000\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot - 4000\right) {|}_{0}^{3}$

$= \left({\left(- 4000 \cdot 3 + 72000\right)}^{\frac{3}{2}} / \left(- 6000\right)\right) - \left({\left(- 4000 \cdot 0 + 72000\right)}^{\frac{3}{2}} / \left(- 6000\right)\right)$

$= {72000}^{\frac{3}{2}} / 6000 - {60000}^{\frac{3}{2}} / 6000$

$= 770.4$

So,

$\overline{v} = \frac{770.4}{3} = 256.8 m {s}^{-} 1$

The average speed is $= 256.8 m {s}^{-} 1$