An object has a mass of $8 k g$ $8 kg$ . The object's kinetic energy uniformly changes from $640 K J$ $640 KJ$ to $320 K J$ $320 KJ$ over $t \in [0, 12 s]$ $t in [0, 12 s]$ . What is the average speed of the object?

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An object has a mass of $8 k g$ $8 kg$ . The object's kinetic energy uniformly changes from $640 K J$ $640 KJ$ to $320 K J$ $320 KJ$ over $t \in [0, 12 s]$ $t in [0, 12 s]$ . What is the average speed of the object?

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Answer 1 · 2017-10-24T05:47:54

The average speed is $= 344.8 m s^{- 1}$ $=344.8ms^-1$

Explanation:

The kinetic energy is

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

The mass is $m = 8 k g$ $m=8kg$

The initial velocity is $= u_{1} m s^{- 1}$ $=u_1ms^-1$

The final velocity is $= u_{2} m s^{- 1}$ $=u_2 ms^-1$

The initial kinetic energy is $\frac{1}{2} m u_{1}^{2} = 640000 J$ $1/2m u_1^2=640000J$

The final kinetic energy is $\frac{1}{2} m u_{2}^{2} = 320000 J$ $1/2m u_2^2=320000J$

Therefore,

$u_{1}^{2} = \frac{2}{8} \cdot 640000 = 160000 m^{2} s^{- 2}$ $u_1^2=2/8*640000=160000m^2s^-2$

and,

$u_{2}^{2} = \frac{2}{8} \cdot 320000 = 80000 m^{2} s^{- 2}$ $u_2^2=2/8*320000=80000m^2s^-2$

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

The points are $(0, 160000)$ $(0,160000)$ and $(12, 80000)$ $(12,80000)$

The equation of the line is

$v^{2} - 160000 = \frac{80000 - 160000}{12} t$ $v^2-160000=(80000-160000)/12t$

$v^{2} = - 6666.7 t + 160000$ $v^2=-6666.7t+160000$

So,

$v = \sqrt{(- 6666.7 t + 160000)}$ $v=sqrt((-6666.7t+160000)$

We need to calculate the average value of $v$ $v$ over $t \in [0, 12]$ $t in [0,12]$

$(12 - 0) ¯ v = \int_{0}^{12} (\sqrt{- 6666.7 t + 160000}) d t$ $(12-0)bar v=int_0^12(sqrt(-6666.7t+160000))dt$

$12 ¯ v = {⎡ ⎣ ⎛ ⎝ \frac{{(- 6666.7 t + 160000)}^{\frac{3}{2}}}{- \frac{3}{2} \cdot 6666.7} ⎞ ⎠ ⎤ ⎦}_{0}^{12}$ $12 barv=[((-6666.7t+160000)^(3/2)/(-3/2*6666.7))]_0^12$

$= ⎛ ⎝ \frac{{(- 6666.7 \cdot 12 + 160000)}^{\frac{3}{2}}}{- 10000} ⎞ ⎠ - ⎛ ⎝ \frac{{(- 1666.7 \cdot 0 + 160000)}^{\frac{3}{2}}}{- 10000} ⎞ ⎠$ $=((-6666.7*12+160000)^(3/2)/(-10000))-((-1666.7*0+160000)^(3/2)/(-10000))$

$= \frac{160000^{\frac{3}{2}}}{10000} - \frac{80000^{\frac{3}{2}}}{10000}$ $=160000^(3/2)/10000-80000^(3/2)/10000$

$= 4137.3$ $=4137.3$

So,

$¯ v = \frac{4137.3}{12} = 344.8 m s^{- 1}$ $barv=4137.3/12=344.8ms^-1$

The average speed is $= 344.8 m s^{- 1}$ $=344.8ms^-1$

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An object has a mass of $8 k g$ $8 kg$ . The object's kinetic energy uniformly changes from $640 K J$ $640 KJ$ to $320 K J$ $320 KJ$ over $t \in [0, 12 s]$ $t in [0, 12 s]$ . What is the average speed of the object?

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An object has a mass of 8 kg8kg8 kg. The object's kinetic energy uniformly changes from 640 KJ640KJ640 KJ to 320 KJ320KJ 320 KJ over t in [0, 12 s]t∈[0,12s]t in [0, 12 s]. What is the average speed of the object?

1 Answer

Explanation:

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An object has a mass of $8 k g$ $8 kg$ . The object's kinetic energy uniformly changes from $640 K J$ $640 KJ$ to $320 K J$ $320 KJ$ over $t \in [0, 12 s]$ $t in [0, 12 s]$ . What is the average speed of the object?