# An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 150 KJ to  16KJ over t in [0,8s]. What is the average speed of the object?

Jun 25, 2017

$168.553416 m {s}^{-} 1$

#### Explanation:

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

150kJ = 150,000J

16kJ = 16,000J

$150 , 000 = \frac{1}{2} \cdot 4 {v}^{2}$
$v = \sqrt{\frac{300 , 000}{4}} = 273.861279 m {s}^{-} 1$

$16 , 000 = \frac{1}{2} \cdot 4 {v}^{2}$
$v = \sqrt{\frac{32 , 000}{4}} = 63.2455532 m {s}^{-} 1$

Proof 1:
$\frac{63.2455532 + 273.861279}{2} = 168.553416 m {s}^{-} 1$

Proof 2:

$s = \frac{t \left(u + v\right)}{2}$ s = distance, t = time, v = final velocity, u = initial velocity.

$\frac{8 \left(273.861279 + 63.2455532\right)}{2} = 1348.42733 m$

"average speed" = ("total distance")/("total time") = 1348.42733/8 = 168.553416ms^-1

Jun 25, 2017

The average speed is $= 197.3 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $= 4 k g$

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

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

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 150000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 16000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{4} \cdot 150000 = 75000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{4} \cdot 16000 = 8000 {m}^{2} {s}^{-} 2$

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

The points are $\left(0 , 75000\right)$ and $\left(8 , 8000\right)$

The equation of the line is

${v}^{2} - 75000 = \frac{8000 - 75000}{8} t$

${v}^{2} = - 8375 t + 75000$

So,

v=sqrt((-8375t+75000)

We need to calculate the average value of $v$ over $t \in \left[0 , 8\right]$

(8-0)bar v=int_0^3sqrt(-8375t+75000))dt

8 barv=[((-8375t+75000)^(3/2)/(-3/2*8375)]_0^8

$= \left({\left(- 8375 \cdot 8 + 75000\right)}^{\frac{3}{2}} / \left(- 12562.5\right)\right) - \left({\left(- 8375 \cdot 0 + 75000\right)}^{\frac{3}{2}} / \left(- 12562.5\right)\right)$

$= {75000}^{\frac{3}{2}} / 12562.5 - {8000}^{\frac{3}{2}} / 12562.5$

$= 1578.03$

So,

$\overline{v} = \frac{1578.03}{8} = 197.3 m {s}^{-} 1$

The average speed is $= 197.3 m {s}^{-} 1$