An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $144 K J$ $144 KJ$ to $120 K J$ $120KJ$ over $t \in [0, 2 s]$ $t in [0, 2 s]$ . 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 $144 K J$ $144 KJ$ to $120 K J$ $120KJ$ over $t \in [0, 2 s]$ $t in [0, 2 s]$ . What is the average speed of the object?

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Answer 1 · 2017-08-14T19:44:03

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

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 $= u_{1} m s^{- 1}$ $=u_1ms^-1$

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

The initial kinetic energy is $\frac{1}{2} m u_{1}^{2} = 144000 J$ $1/2m u_1^2=144000J$

The final kinetic energy is $\frac{1}{2} m u_{2}^{2} = 120000 J$ $1/2m u_2^2=120000J$

Therefore,

$u_{1}^{2} = \frac{2}{4} \cdot 144000 = 72000 m^{2} s^{- 2}$ $u_1^2=2/4*144000=72000m^2s^-2$

and,

$u_{2}^{2} = \frac{2}{4} \cdot 120000 = 60000 m^{2} s^{- 2}$ $u_2^2=2/4*120000=60000m^2s^-2$

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

The points are $(0, 72000)$ $(0,72000)$ and $(2, 60000)$ $(2,60000)$

The equation of the line is

$v^{2} - 72000 = \frac{60000 - 72000}{2} t$ $v^2-72000=(60000-72000)/2t$

$v^{2} = - 6000 t + 72000$ $v^2=-6000t+72000$

So,

$v = \sqrt{(- 6000 t + 72000)}$ $v=sqrt((-6000t+72000)$

We need to calculate the average value of $v$ $v$ over $t \in [0, 2]$ $t in [0,2]$

$(2 - 0) ¯ v = \int_{0}^{2} (\sqrt{- 6000 t + 72000}) d t$ $(2-0)bar v=int_0^2(sqrt(-6000t+72000))dt$

$2 ¯ v = {⎡ ⎣ ⎛ ⎝ \frac{{(- 6000 t + 72000)}^{\frac{3}{2}}}{\frac{3}{2} \cdot - 6000} ⎞ ⎠ ⎤ ⎦}_{0}^{2}$ $2 barv=[((-6000t+72000)^(3/2)/(3/2*-6000))]_0^2$

$= ⎛ ⎝ \frac{{(- 6000 \cdot 2 + 72000)}^{\frac{3}{2}}}{- 9000} ⎞ ⎠ - ⎛ ⎝ \frac{{(- 6000 \cdot 0 + 72000)}^{\frac{3}{2}}}{- 9000} ⎞ ⎠$ $=((-6000*2+72000)^(3/2)/(-9000))-((-6000*0+72000)^(3/2)/(-9000))$

$= \frac{72000^{\frac{3}{2}}}{9000} - \frac{60000^{\frac{3}{2}}}{9000}$ $=72000^(3/2)/9000-60000^(3/2)/9000$

$= 513.6$ $=513.6$

So,

$¯ v = \frac{513.6}{2} = 256.8 m s^{- 1}$ $barv=513.6/2=256.8ms^-1$

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

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An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $144 K J$ $144 KJ$ to $120 K J$ $120KJ$ over $t \in [0, 2 s]$ $t in [0, 2 s]$ . 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 144 KJ144KJ144 KJ to 120KJ120KJ 120KJ over t in [0, 2 s]t∈[0,2s]t in [0, 2 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 $144 K J$ $144 KJ$ to $120 K J$ $120KJ$ over $t \in [0, 2 s]$ $t in [0, 2 s]$ . What is the average speed of the object?