An object has a mass of $3 k g$ $3 kg$ . The object's kinetic energy uniformly changes from $48 K J$ $48 KJ$ to $32 K J$ $32KJ$ over $t \in [0, 12 s]$ $t in [0,12s]$ . What is the average speed of the object?

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An object has a mass of $3 k g$ $3 kg$ . The object's kinetic energy uniformly changes from $48 K J$ $48 KJ$ to $32 K J$ $32KJ$ over $t \in [0, 12 s]$ $t in [0,12s]$ . What is the average speed of the object?

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Answer 1 · 2017-09-23T05:46:04

The average speed is $= 163.0 m s^{- 1}$ $=163.0ms^-1$

Explanation:

The kinetic energy is

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

The mass is $m = 3 k g$ $m=3kg$

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 $\frac{1}{2} m u_{1}^{2} = 48000 J$ $1/2m u_1^2=48000J$

The final kinetic energy is $\frac{1}{2} m u_{2}^{2} = 32000 J$ $1/2m u_2^2=32000J$

Therefore,

$u_{1}^{2} = \frac{2}{3} \cdot 48000 = 32000 m^{2} s^{- 2}$ $u_1^2=2/3*48000=32000m^2s^-2$

and,

$u_{2}^{2} = \frac{2}{3} \cdot 32000 = 21333.3 m^{2} s^{- 2}$ $u_2^2=2/3*32000=21333.3m^2s^-2$

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

The points are $(0, 32000)$ $(0,32000)$ and $(12, 21333.3)$ $(12,21333.3)$

The equation of the line is

$v^{2} - 32000 = \frac{21333.3 - 32000}{12} t$ $v^2-32000=(21333.3-32000)/12t$

$v^{2} = - 888.9 t + 32000$ $v^2=-888.9t+32000$

So,

$v = \sqrt{(- 888.9 t + 32000)}$ $v=sqrt((-888.9t+32000)$

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

$(12 - 0) ¯ v = \int_{0}^{12} (\sqrt{- 888.9 t + 32000}) d t$ $(12-0)bar v=int_0^12(sqrt(-888.9t+32000))dt$

$12 ¯ v = {⎡ ⎣ ⎛ ⎝ \frac{{(- 888.9 t + 32000)}^{\frac{3}{2}}}{- \frac{3}{2} \cdot 888.9} ⎞ ⎠ ⎤ ⎦}_{0}^{12}$ $12 barv=[((-888.9t+32000)^(3/2)/(-3/2*888.9))]_0^12$

$= ⎛ ⎝ \frac{{(- 888.9 \cdot 12 + 32000)}^{\frac{3}{2}}}{- 1333.3} ⎞ ⎠ - ⎛ ⎝ \frac{{(- 888.9 \cdot 0 + 32000)}^{\frac{3}{2}}}{- 1333.3} ⎞ ⎠$ $=((-888.9*12+32000)^(3/2)/(-1333.3))-((-888.9*0+32000)^(3/2)/(-1333.3))$

$= \frac{32000^{\frac{3}{2}}}{1333.3} - \frac{{21333.3}^{\frac{3}{2}}}{1333.3}$ $=32000^(3/2)/1333.3-21333.3^(3/2)/1333.3$

$= 1956.4$ $=1956.4$

So,

$¯ v = \frac{1956.4}{12} = 163.0 m s^{- 1}$ $barv=1956.4/12=163.0ms^-1$

The average speed is $= 163.0 m s^{- 1}$ $=163.0ms^-1$

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An object has a mass of $3 k g$ $3 kg$ . The object's kinetic energy uniformly changes from $48 K J$ $48 KJ$ to $32 K J$ $32KJ$ over $t \in [0, 12 s]$ $t in [0,12s]$ . What is the average speed of the object?

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Explanation:

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An object has a mass of 3kg3 kg. The object's kinetic energy uniformly changes from 48KJ48 KJ to 32KJ 32KJ over t∈[0,12s]t in [0,12s]. What is the average speed of the object?

1 Answer

Explanation:

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An object has a mass of $3 k g$ $3 kg$ . The object's kinetic energy uniformly changes from $48 K J$ $48 KJ$ to $32 K J$ $32KJ$ over $t \in [0, 12 s]$ $t in [0,12s]$ . What is the average speed of the object?