An object has a mass of $2 k g$ $2 kg$ . The object's kinetic energy uniformly changes from $35 K J$ $35 KJ$ to $18 K J$ $18 KJ$ 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 $35 K J$ $35 KJ$ to $18 K J$ $18 KJ$ 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 · 2018-01-13T10:25:00

The average speed is $= 162.1 m s^{- 1}$ $=162.1ms^-1$

Explanation:

The kinetic energy is

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

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

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 $\frac{1}{2} m u_{1}^{2} = 35000 J$ $1/2m u_1^2=35000J$

The final kinetic energy is $\frac{1}{2} m u_{2}^{2} = 18000 J$ $1/2m u_2^2=18000J$

Therefore,

$u_{1}^{2} = \frac{2}{2} \cdot 35000 = 35000 m^{2} s^{- 2}$ $u_1^2=2/2*35000=35000m^2s^-2$

and,

$u_{2}^{2} = \frac{2}{2} \cdot 18000 = 18000 m^{2} s^{- 2}$ $u_2^2=2/2*18000=18000m^2s^-2$

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

The points are $(0, 35000)$ $(0,35000)$ and $(15, 18000)$ $(15,18000)$

The equation of the line is

$v^{2} - 35000 = \frac{18000 - 35000}{15} t$ $v^2-35000=(18000-35000)/15t$

$v^{2} = - 1133.3 t + 35000$ $v^2=-1133.3t+35000$

So,

$v = \sqrt{- 1133.3 t + 35000}$ $v=sqrt(-1133.3t+35000)$

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

$(15 - 0) ¯ v = \int_{0}^{15} (\sqrt{- 1133.3 t + 35000}) d t$ $(15-0)bar v=int_0^15(sqrt(-1133.3t+35000))dt$

$15 ¯ v = \frac{{(- 1133.3 t + 35000)}^{\frac{3}{2}}}{\frac{3}{2} \cdot - 1133.3} {∣ ∣ ∣ ∣}_{0}^{15}$ $15 barv= (-1133.3t+35000)^(3/2)/(3/2*-1133.3)| _( 0) ^ (15)$

$= ⎛ ⎝ \frac{{(- 1133.3 \cdot 15 + 35000)}^{\frac{3}{2}}}{- 1700} ⎞ ⎠ - ⎛ ⎝ \frac{{(- 1133.3 \cdot 0 + 35000)}^{\frac{3}{2}}}{- 1700} ⎞ ⎠$ $=((-1133.3*15+35000)^(3/2)/(-1700))-((-1133.3*0+35000)^(3/2)/(-1700))$

$= \frac{35000^{\frac{3}{2}}}{1700} - \frac{18000^{\frac{3}{2}}}{1700}$ $=35000^(3/2)/1700-18000^(3/2)/1700$

$= 2431.1$ $=2431.1$

So,

$¯ v = \frac{2431.1}{15} = 162.1 m s^{- 1}$ $barv=2431.1/15=162.1ms^-1$

The average speed is $= 162.1 m s^{- 1}$ $=162.1ms^-1$

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An object has a mass of $2 k g$ $2 kg$ . The object's kinetic energy uniformly changes from $35 K J$ $35 KJ$ to $18 K J$ $18 KJ$ 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 $35 K J$ $35 KJ$ to $18 K J$ $18 KJ$ over $t \in [0, 15 s]$ $t in [0, 15 s]$ . What is the average speed of the object?