What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $7 \frac{m}{s}$ $7 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = t^{2} - 2 t$ $a(t) =t^2-2t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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Answer 1 · 2017-12-05T14:45:26

The average speed is $= 8.27 m s^{- 1}$ $=8.27ms^-1$

The speed is the integral of the acceleration

$a (t) = t^{2} - 2 t$ $a(t)=t^2-2t$

Then,

$v (t) = \int (t^{2} - 2 t) d t = \frac{1}{3} t^{3} - \frac{2}{2} t^{2} + C$ $v(t)=int(t^2-2t)dt=1/3t^3-2/2t^2+C$

Plugging in the initial conditions at $t = 0$ $t=0$

$v (0) = 7 = 0 - 0 + C$ $v(0)=7=0-0+C$

$C = 7$ $C=7$

Therefore,

$v (t) = \frac{1}{3} t^{3} - t^{2} + 7$ $v(t)=1/3t^3-t^2+7$ in the interval $[0, 4]$ $[0,4]$

After that $t > 4$ $t>4$ , assume that the speed is constant

$v (4) = \frac{1}{3} \cdot 4^{3} - 4^{2} + 8 = \frac{64}{3} - 8 = 13.33 m s^{- 1}$ $v(4)=1/3*4^3-4^2+8=64/3-8=13.33ms^-1$

The average speed on the interval $[0, 5]$ $[0,5]$ is

$5 ¯ v = \int_{0}^{4} (\frac{1}{3} t^{3} - t^{2} + 7) d t + 13.33 \cdot 1$ $5barv=int_0^4(1/3t^3-t^2+7)dt+13.33*1$

$= {[\frac{1}{12} t^{4} - \frac{1}{3} t^{3} + 7 t]}_{0}^{4} + 13.33$ $=[1/12t^4-1/3t^3+7t]_0^4+13.33$

$= (\frac{1}{12} \cdot 4^{4} - \frac{4^{3}}{3} + 7 \cdot 4) + 13.33$ $=(1/12*4^4-4^3/3+7*4)+13.33$

$= 41.33$ $=41.33$

The average speed is

$¯ v = \frac{41.33}{5} = 8.27 m s^{- 1}$ $barv=41.33/5=8.27ms^-1$

What is the average speed, on t∈[0,5]t in [0,5], of an object that is moving at 7ms7 m/s at t=0t=0 and accelerates at a rate of a(t)=t2−2ta(t) =t^2-2t on t∈[0,4]t in [0,4]?