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 $a (t) = 2 t^{2} + 1$ $a(t) =2t^2+1$ on $t \in [0, 2]$ $t in [0,2]$ ?

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Answer 1 · 2017-04-15T06:57:45

The average speed is $= 8.33 m s^{- 1}$ $=8.33ms^-1$

The velocity is the integral of the acceleration

$a (t) = 2 t^{2} + 1$ $a(t)=2t^2+1$

$v (t) = \int (2 t^{2} + 1) d t$ $v(t)=int(2t^2+1)dt$

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

Plugging in the initial conditions

$v (0) = 3$ $v(0)=3$

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

So,

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

$v (2) = \frac{16}{3} + 5 = \frac{31}{3}$ $v(2)=16/3+5=31/3$

The average speed is

$(5 - 0) ¯ v = \int_{0}^{2} (\frac{2}{3} t^{3} + t + 3) d t + (\frac{31}{3} \cdot 3)$ $(5-0)barv=int_0^2(2/3t^3+t+3)dt+(31/3*3)$

$5 ¯ v = {[\frac{2}{3} \cdot \frac{1}{4} \cdot t^{4} + \frac{1}{2} t^{2} + 3 t]}_{0}^{4} + 31$ $5barv=[2/3*1/4*t^4+1/2t^2+3t]_0^2+31$

$= \frac{8}{3} + 2 + 6 + 31 = \frac{125}{3}$ $=8/3+2+6+31=125/3$

$¯ v = \frac{1}{5} \cdot \frac{125}{3} = \frac{25}{3} = 8.33 m s^{- 1}$ $barv=1/5*125/3=25/3=8.33ms^-1$

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+1a(t)=2t2+1a(t) =2t^2+1 on t in [0,2]t∈[0,2]t in [0,2]?