An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $64 K J$ $64 KJ$ to $180 K J$ $180 KJ$ 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 $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $64 K J$ $64 KJ$ to $180 K J$ $180 KJ$ 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 · 2017-06-15T17:37:12

The average speed is $= 244.5 m s^{- 1}$ $=244.5ms^-1$

Explanation:

The kinetic energy is

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

The mass is $= 4 k g$ $=4kg$

The initial velocity is $= u_{1}$ $=u_1$

The initial kinetic energy is $\frac{1}{2} m u_{1}^{2} = 64000 J$ $1/2m u_1^2=64000J$

The final velocity is $= u_{2}$ $=u_2$

The final kinetic energy is $\frac{1}{2} m u_{2}^{2} = 180000 J$ $1/2m u_2^2=180000J$

Therefore,

$u_{1}^{2} = \frac{2}{4} \cdot 64000 = 32000 m^{2} s^{- 2}$ $u_1^2=2/4*64000=32000m^2s^-2$

and,

$u_{2}^{2} = \frac{2}{4} \cdot 180000 = 90000 m^{2} s^{- 2}$ $u_2^2=2/4*180000=90000m^2s^-2$

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

The points are $(0, 32000)$ $(0,32000)$ and $(12, 90000)$ $(12,90000)$

The equation of the line is

$v^{2} - 32000 = \frac{90000 - 32000}{12} t$ $v^2-32000=(90000-32000)/12t$

$v^{2} = 4833.3 t + 32000$ $v^2=4833.3t+32000$

So,

$v = \sqrt{(4833.3 t + 32000)}$ $v=sqrt((4833.3t+32000)$

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

$(12 - 0) ¯ v = \int_{0}^{12} \sqrt{(4833.3 t + 32000)} d t$ $(12-0)bar v=int_0^12sqrt((4833.3t+32000))dt$

$12 ¯ v = ⎡ ⎢ ⎣ {⎛ ⎝ \frac{{(4833.3 t + 32000)}^{\frac{3}{2}}}{\frac{3}{2} \cdot 4833.3} ⎤ ⎦}_{0}^{12}$ $12 barv=[((4833.3t+32000)^(3/2)/(3/2*4833.3)]_0^12$

$= ⎛ ⎝ \frac{{(4833.3 \cdot 12 + 32000)}^{\frac{3}{2}}}{7250} ⎞ ⎠ - ⎛ ⎝ \frac{{(4833.3 \cdot 0 + 32000)}^{\frac{3}{2}}}{7250} ⎞ ⎠$ $=((4833.3*12+32000)^(3/2)/(7250))-((4833.3*0+32000)^(3/2)/(7250))$

$= \frac{90000^{\frac{3}{2}}}{7250} - \frac{32000^{\frac{3}{2}}}{7250}$ $=90000^(3/2)/7250-32000^(3/2)/7250$

$= 2934.6$ $=2934.6$

So,

$¯ v = \frac{2934.6}{12} = 244.5 m s^{- 1}$ $barv=2934.6/12=244.5ms^-1$

The average speed is $= 244.5 m s^{- 1}$ $=244.5ms^-1$

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An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $64 K J$ $64 KJ$ to $180 K J$ $180 KJ$ 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 4 kg4kg4 kg. The object's kinetic energy uniformly changes from 64 KJ64KJ64 KJ to 180 KJ180KJ 180 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 $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $64 K J$ $64 KJ$ to $180 K J$ $180 KJ$ over $t \in [0, 12 s]$ $t in [0, 12 s]$ . What is the average speed of the object?