What is the average speed of an object that is moving at $9 \frac{m}{s}$ $9 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 3 - t$ $a(t) =3-t$ on $t \in [0, 5]$ $t in [0,5]$ ?

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Answer 1 · 2018-05-07T14:35:01

The average speed is $= 12.33 m s^{- 1}$ $=12.33ms^-1$

$a (t) = 3 - t$ $a(t)=3-t$

The velocity is the integral of the acceleration

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

Plugging in the initial conditions

$v (0) = 9$ $v(0)=9$

Therefore,

$v (0) = 0 - 0 + C = 9$ $v(0)=0-0+C=9$

$C = 9$ $C=9$

$v (t) = 3 t - \frac{1}{2} t^{2} + 9$ $v(t)=3t-1/2t^2+9$

The average speed on $t \in [0, 5]$ $t in [0,5]$ is

$(5 - 0) ¯ v = \int_{0}^{5} (3 t - \frac{1}{2} t^{2} + 9) d t$ $(5-0)barv=int_0^5(3t-1/2t^2+9)dt$

$(5) ¯ v = {[\frac{3}{2} t^{2} - \frac{1}{6} t^{3} + 9 t]}_{0}^{2}$ $(5)barv=[3/2t^2-1/6t^3+9t]_0^5$

$= (\frac{75}{2} - \frac{125}{6} + 45) - (0)$ $=(75/2-125/6+45)-(0)$

$= \frac{185}{3}$ $=185/3$

$¯ v = \frac{185}{15} = 12.33 m s^{- 1}$ $barv=185/15=12.33ms^-1$

What is the average speed of an object that is moving at 9ms9 m/s at t=0t=0 and accelerates at a rate of a(t)=3−ta(t) =3-t on t∈[0,5]t in [0,5]?