An object has a mass of $2 k g$ $2 kg$ . The object's kinetic energy uniformly changes from $64 K J$ $64 KJ$ to $38 K J$ $38 KJ$ over $t \in [0, 15 s]$ $t in [0, 15 s]$ . What is the average speed of the object?

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An object has a mass of $2 k g$ $2 kg$ . The object's kinetic energy uniformly changes from $64 K J$ $64 KJ$ to $38 K J$ $38 KJ$ over $t \in [0, 15 s]$ $t in [0, 15 s]$ . What is the average speed of the object?

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Answer 1 · 2017-09-12T05:42:14

The average speed is $= 225.2 m s^{- 1}$ $=225.2ms^-1$

Explanation:

The kinetic energy is

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

The mass is $= 2 k g$ $=2kg$

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 $\frac{1}{2} m u_{1}^{2} = 64000 J$ $1/2m u_1^2=64000J$

The final kinetic energy is $\frac{1}{2} m u_{2}^{2} = 38000 J$ $1/2m u_2^2=38000J$

Therefore,

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

and,

$u_{2}^{2} = \frac{2}{2} \cdot 38000 = 38000 m^{2} s^{- 2}$ $u_2^2=2/2*38000=38000m^2s^-2$

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

The points are $(0, 64000)$ $(0,64000)$ and $(15, 38000)$ $(15,38000)$

The equation of the line is

$v^{2} - 64000 = \frac{38000 - 64000}{15} t$ $v^2-64000=(38000-64000)/15t$

$v^{2} = - 17333 t + 64000$ $v^2=-17333t+64000$

So,

$v = \sqrt{(- 1733.3 t + 64000)}$ $v=sqrt((-1733.3t+64000)$

We need to calculate the average value of $v$ $v$ over $t \in [0, 15]$ $t in [0,15]$

$(15 - 0) ¯ v = \int_{0}^{15} (\sqrt{- 1733.3 t + 64000}) d t$ $(15-0)bar v=int_0^15(sqrt(-1733.3t+64000))dt$

$15 ¯ v = {⎡ ⎣ ⎛ ⎝ \frac{{(- 1733.3 t + 64000)}^{\frac{3}{2}}}{- \frac{3}{2} \cdot 1733.3} ⎞ ⎠ ⎤ ⎦}_{0}^{15}$ $15 barv=[((-1733.3t+64000)^(3/2)/(-3/2*1733.3))]_0^15$

$= ⎛ ⎝ \frac{{(- 1733.3 \cdot 15 + 64000)}^{\frac{3}{2}}}{- 2600} ⎞ ⎠ - ⎛ ⎝ \frac{{(- 1733.3 \cdot 0 + 64000)}^{\frac{3}{2}}}{- 2600} ⎞ ⎠$ $=((-1733.3*15+64000)^(3/2)/(-2600))-((-1733.3*0+64000)^(3/2)/(-2600))$

$= \frac{64000^{\frac{3}{2}}}{2600} - \frac{38000^{\frac{3}{2}}}{2600}$ $=64000^(3/2)/2600-38000^(3/2)/2600$

$= 3378.2$ $=3378.2$

So,

$¯ v = \frac{3378.2}{15} = 225.2 m s^{- 1}$ $barv=3378.2/15=225.2ms^-1$

The average speed is $= 225.2 m s^{- 1}$ $=225.2ms^-1$

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An object has a mass of $2 k g$ $2 kg$ . The object's kinetic energy uniformly changes from $64 K J$ $64 KJ$ to $38 K J$ $38 KJ$ over $t \in [0, 15 s]$ $t in [0, 15 s]$ . What is the average speed of the object?

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1 Answer

Explanation:

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An object has a mass of $2 k g$ $2 kg$ . The object's kinetic energy uniformly changes from $64 K J$ $64 KJ$ to $38 K J$ $38 KJ$ over $t \in [0, 15 s]$ $t in [0, 15 s]$ . What is the average speed of the object?