An object has a mass of $6 k g$ $6 kg$ . The object's kinetic energy uniformly changes from $48 K J$ $48 KJ$ to $15 K J$ $15 KJ$ over $t \in [0, 8 s]$ $t in [0, 8 s]$ . What is the average speed of the object?

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An object has a mass of $6 k g$ $6 kg$ . The object's kinetic energy uniformly changes from $48 K J$ $48 KJ$ to $15 K J$ $15 KJ$ over $t \in [0, 8 s]$ $t in [0, 8 s]$ . What is the average speed of the object?

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Answer 1 · 2017-05-15T12:15:43

The average speed is $= 101.23 m s^{- 1}$ $=101.23ms^-1$

Explanation:

The kinetic energy is

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

mass is $= 6 k g$ $=6kg$

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

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

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

$\frac{1}{2} m u_{2}^{2} = 15000 J$ $1/2m u_2^2=15000J$

Therefore,

$u_{1}^{2} = \frac{2}{6} \cdot 48000 = 16000 m^{2} s^{- 2}$ $u_1^2=2/6*48000=16000m^2s^-2$

and,

$u_{2}^{2} = \frac{2}{6} \cdot 15000 = 5000 m^{2} s^{- 2}$ $u_2^2=2/6*15000=5000m^2s^-2$

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

The points are $(0, 16000)$ $(0,16000)$ and $(8, 5000)$ $(8,5000)$

The equation of the line is

$v^{2} - 16000 = \frac{5000 - 16000}{8} t$ $v^2-16000=(5000-16000)/8t$

$v^{2} = - 1375 t + 16000$ $v^2=-1375t+16000$

So,

$v = \sqrt{(- 1375 t + 16000)}$ $v=sqrt((-1375t+16000)$

We need to calculate the average value of $v$ $v$ over $t \in [0, 8]$ $t in [0,8]$

$(8 - 0) ¯ v = \int_{0}^{8} \sqrt{(- 1375 t + 16000)} d t$ $(8-0)bar v=int_0^8sqrt((-1375t+16000))dt$

$8 ¯ v = ⎡ ⎢ ⎣ {⎛ ⎝ \frac{{(- 1375 t + 16000)}^{\frac{3}{2}}}{- \frac{3}{2} \cdot 1375} ⎤ ⎦}_{0}^{8}$ $8 barv=[((-1375t+16000)^(3/2)/(-3/2*1375)]_0^8$

$= ⎛ ⎝ \frac{{(- 1375 \cdot 8 + 16000)}^{\frac{3}{2}}}{- 2062.5} ⎞ ⎠ - ⎛ ⎝ \frac{{(- 1375 \cdot 0 + 16000)}^{\frac{3}{2}}}{- 2062.5} ⎞ ⎠$ $=((-1375*8+16000)^(3/2)/(-2062.5))-((-1375*0+16000)^(3/2)/(-2062.5))$

$= \frac{16000^{\frac{3}{2}}}{2062.5} - \frac{5000^{\frac{3}{2}}}{2062.5}$ $=16000^(3/2)/2062.5-5000^(3/2)/2062.5$

$= 809.84$ $=809.84$

So,

$¯ v = \frac{809.84}{8} = 101.23 m s^{- 1}$ $barv=809.84/8=101.23ms^-1$

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An object has a mass of $6 k g$ $6 kg$ . The object's kinetic energy uniformly changes from $48 K J$ $48 KJ$ to $15 K J$ $15 KJ$ over $t \in [0, 8 s]$ $t in [0, 8 s]$ . What is the average speed of the object?

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An object has a mass of 6 kg6kg6 kg. The object's kinetic energy uniformly changes from 48 KJ48KJ48 KJ to 15 KJ15KJ 15 KJ over t in [0, 8 s]t∈[0,8s]t in [0, 8 s]. What is the average speed of the object?

1 Answer

Explanation:

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An object has a mass of $6 k g$ $6 kg$ . The object's kinetic energy uniformly changes from $48 K J$ $48 KJ$ to $15 K J$ $15 KJ$ over $t \in [0, 8 s]$ $t in [0, 8 s]$ . What is the average speed of the object?