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

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Answer 1 · 2018-03-02T09:26:12

The average speed is $= 12.83 m s^{- 1}$ $=12.83ms^-1$

The speed is the integral of the acceleration

$a (t) = 2 t + 1$ $a(t)=2t+1$

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

$= t^{2} + t + C$ $=t^2+t+C$

Plugging in the initial conditions

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

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

$\Rightarrow$ $=>$ , $C = 9$ $C=9$

Therefore,

$v (t) = t^{2} + t + 9$ $v(t)=t^2+t+9$

The average speed is

$(2 -) ¯ v = \int_{1}^{2} (t^{2} + t + 9) d t$ $(2-)barv=int_1^2(t^2+t+9)dt$

$¯ v = {[\frac{t^{3}}{3} + \frac{t^{2}}{2} + 9 t]}_{1}^{2}$ $barv=[t^3/3+t^2/2+9t]_1^2$

$= (\frac{8}{3} + 2 + 18) - (\frac{1}{3} + \frac{1}{2} + 9)$ $=(8/3+2+18)-(1/3+1/2+9)$

$= \frac{7}{3} + 11 - \frac{1}{2}$ $=7/3+11-1/2$

$= 12.83 m s^{- 1}$ $=12.83ms^-1$

What is the average speed of an object that is moving at 9 m/s9ms9 m/s at t=0t=0t=0 and accelerates at a rate of a(t) =2t+1a(t)=2t+1a(t) =2t+1 on t in [1,2]t∈[1,2]t in [1,2]?