What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $3 \frac{m}{s}$ $3 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 2 t^{2} - 2$ $a(t) =2t^2-2$ on $t \in [0, 2]$ $t in [0,2]$ ?

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Answer 1 · 2017-05-02T06:47:55

The average speed is $= 3.53 m s^{- 1}$ $=3.53ms^-1$

The speed is the integral of the acceleration

$v (t) = \int a (t) d t$ $v(t)=inta(t)dt$

$= \int (2 t^{2} - 2) d t$ $=int(2t^2-2)dt$

$= \frac{2}{3} t^{3} - 2 t + C$ $=2/3t^3-2t+C$

Plugging in the initial conditions

$v (0) = 0 - 0 + C = 3$ $v(0)=0-0+C=3$

Therefore,

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

$v (2) = \frac{2}{3} \cdot 2^{3} - 2 \cdot 2 + 3 = \frac{13}{3}$ $v(2)=2/3*2^3-2*2+3=13/3$

The average value is

$(5 - 0) ¯ v = \int_{0}^{2} (\frac{2}{3} t^{3} - 2 t + 3) d t + \frac{13}{3} \cdot 3$ $(5-0)barv=int_0^2(2/3t^3-2t+3)dt+13/3*3$

$5 ¯ v = {[\frac{2}{3} \cdot \frac{1}{4} \cdot t^{4} - \frac{2}{2} t^{2} + 3 t]}_{0}^{4} + 13$ $5barv=[2/3*1/4*t^4-2/2t^2+3t]_0^2+13$

$= (\frac{8}{3} - 4 + 6) - (0) + 13$ $=(8/3-4+6)-(0)+13$

$= \frac{14}{3} + 13 = \frac{53}{3}$ $=14/3+13=53/3$

$¯ v = \frac{53}{15} = 3.53 m s^{- 1}$ $barv=53/15=3.53ms^-1$

What is the average speed, on t in [0,5]t∈[0,5]t in [0,5], of an object that is moving at 3 m/s3ms3 m/s at t=0t=0t=0 and accelerates at a rate of a(t) =2t^2-2a(t)=2t2−2a(t) =2t^2-2 on t in [0,2]t∈[0,2]t in [0,2]?