An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $254 K J$ $254 KJ$ to $24 K J$ $24 KJ$ over $t \in [0, 5 s]$ $t in [0, 5 s]$ . 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 $254 K J$ $254 KJ$ to $24 K J$ $24 KJ$ over $t \in [0, 5 s]$ $t in [0, 5 s]$ . What is the average speed of the object?

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Answer 1 · 2017-04-13T06:54:57

The average speed is $= 254.75 m s^{- 1}$ $=254.75ms^-1$

Explanation:

The kinetic energy is

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

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

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

$\frac{1}{2} m u_{1}^{2} = 254000 J$ $1/2m u_1^2=254000J$

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

$\frac{1}{2} m u_{2}^{2} = 24000 J$ $1/2m u_2^2=24000J$

Therefore,

$u_{1}^{2} = \frac{2}{4} \cdot 254000 = 127000 m^{2} s^{- 2}$ $u_1^2=2/4*254000=127000m^2s^-2$

and,

$u_{2}^{2} = \frac{2}{4} \cdot 24000 = 12000 m^{2} s^{- 2}$ $u_2^2=2/4*24000=12000m^2s^-2$

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

The points are $(0, 127000)$ $(0,127000)$ and $(5, 12000)$ $(5,12000)$

The equation of the line is

$v^{2} - 127000 = \frac{12000 - 127000}{5} t$ $v^2-127000=(12000-127000)/5t$

$v^{2} = - 23000 t + 127000$ $v^2=-23000t+127000$

So,

$v = \sqrt{(- 23000 t + 127000)}$ $v=sqrt((-23000t+127000)$

We need to calculate the average value of $v$ $v$ over $t \in [0, 5]$ $t in [0,5]$

$(5 - 0) ¯ v = \int_{0}^{5} \sqrt{(- 23000 t + 127000)} d t$ $(5-0)bar v=int_0^5sqrt((-23000t+127000))dt$

$5 ¯ v = ⎡ ⎢ ⎣ {⎛ ⎝ \frac{{(- 23000 t + 127000)}^{\frac{3}{2}}}{- \frac{3}{2} \cdot 23000} ⎤ ⎦}_{0}^{5}$ $5 barv=[((-23000t+127000)^(3/2)/(-3/2*23000)]_0^5$

$= ⎛ ⎝ \frac{{(- 23000 \cdot 5 + 127000)}^{\frac{3}{2}}}{- 34500} ⎞ ⎠ - ⎛ ⎝ \frac{{(- 23000 \cdot 0 + 127000)}^{\frac{3}{2}}}{- 34500} ⎞ ⎠$ $=((-23000*5+127000)^(3/2)/(-34500))-((-23000*0+127000)^(3/2)/(-34500))$

$= \frac{127000^{\frac{3}{2}}}{34500} - \frac{12000^{\frac{3}{2}}}{34500}$ $=127000^(3/2)/34500-12000^(3/2)/34500$

$= 1273.75$ $=1273.75$

So,

$¯ v = \frac{1273.75}{5} = 254.75 m s^{- 1}$ $barv=1273.75/5=254.75ms^-1$

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An object has a mass of $4 k g$ $4 kg$ . The object's kinetic energy uniformly changes from $254 K J$ $254 KJ$ to $24 K J$ $24 KJ$ over $t \in [0, 5 s]$ $t in [0, 5 s]$ . 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 254 KJ254KJ254 KJ to 24 KJ24KJ 24 KJ over t in [0, 5 s]t∈[0,5s]t in [0, 5 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 $254 K J$ $254 KJ$ to $24 K J$ $24 KJ$ over $t \in [0, 5 s]$ $t in [0, 5 s]$ . What is the average speed of the object?