An object has a mass of $6 k g$ $6 kg$ . The object's kinetic energy uniformly changes from $120 K J$ $120 KJ$ to $720 K J$ $720 KJ$ over $t \in [0, 3 s]$ $t in [0, 3 s]$ . 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 $120 K J$ $120 KJ$ to $720 K J$ $720 KJ$ over $t \in [0, 3 s]$ $t in [0, 3 s]$ . What is the average speed of the object?

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Answer 1 · 2017-12-12T10:57:36

The average speed is $= 365.3 m s^{- 1}$ $=365.3ms^-1$

Explanation:

The kinetic energy is

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

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

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 $\frac{1}{2} m u_{1}^{2} = 120000 J$ $1/2m u_1^2=120000J$

The final kinetic energy is $\frac{1}{2} m u_{2}^{2} = 720000 J$ $1/2m u_2^2=720000J$

Therefore,

$u_{1}^{2} = \frac{2}{6} \cdot 120000 = 40000 m^{2} s^{- 2}$ $u_1^2=2/6*120000=40000m^2s^-2$

and,

$u_{2}^{2} = \frac{2}{6} \cdot 720000 = 240000 m^{2} s^{- 2}$ $u_2^2=2/6*720000=240000m^2s^-2$

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

The points are $(0, 40000)$ $(0,40000)$ and $(3, 240000)$ $(3,240000)$

The equation of the line is

$v^{2} - 40000 = \frac{240000 - 40000}{3} t$ $v^2-40000=(240000-40000)/3t$

$v^{2} = 66666.7 t + 40000$ $v^2=66666.7t+40000$

So,

$v = \sqrt{66666.7 t + 40000}$ $v=sqrt(66666.7t+40000)$

We need to calculate the average value of $v$ $v$ over $t \in [0, 3]$ $t in [0,3]$

$(3 - 0) ¯ v = \int_{0}^{3} (\sqrt{66666.7 t + 40000}) d t$ $(3-0)bar v=int_0^3(sqrt(66666.7t+40000))dt$

$3 ¯ v = \frac{{[66666.7 t + 40000)}^{\frac{3}{2}}}{\frac{3}{2} \cdot 66666.7} {∣ ∣ ∣ ∣}_{0}^{3}$ $3 barv= [ 66666.7t+40000)^(3/2)/(3/2*66666.7)| _( 0) ^ (3)$

$= ⎛ ⎝ \frac{{(66666.7 \cdot 3 + 40000)}^{\frac{3}{2}}}{100000} ⎞ ⎠ - ⎛ ⎝ \frac{{(66666.7 \cdot 0 + 40000)}^{\frac{3}{2}}}{100000} ⎞ ⎠$ $=((66666.7*3+40000)^(3/2)/(100000))-((66666.7*0+40000)^(3/2)/(100000))$

$= \frac{240000^{\frac{3}{2}}}{100000} - \frac{40000^{\frac{3}{2}}}{100000}$ $=240000^(3/2)/100000-40000^(3/2)/100000$

$= 1095.76$ $=1095.76$

So,

$¯ v = \frac{1095.76}{3} = 365.3 m s^{- 1}$ $barv=1095.76/3=365.3ms^-1$

The average speed is $= 365.3 m s^{- 1}$ $=365.3ms^-1$

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An object has a mass of $6 k g$ $6 kg$ . The object's kinetic energy uniformly changes from $120 K J$ $120 KJ$ to $720 K J$ $720 KJ$ over $t \in [0, 3 s]$ $t in [0, 3 s]$ . 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 120 KJ120KJ120 KJ to 720 KJ720KJ 720 KJ over t in [0, 3 s]t∈[0,3s]t in [0, 3 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 $120 K J$ $120 KJ$ to $720 K J$ $720 KJ$ over $t \in [0, 3 s]$ $t in [0, 3 s]$ . What is the average speed of the object?