What is the average speed of an object that is moving at $8 \frac{m}{s}$ $8 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 2 - t^{2}$ $a(t) =2-t^2$ on $t \in [0, 2]$ $t in [0,2]$ ?

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Answer 1 · 2017-04-08T06:27:33

The average speed is $= 9.33 m s^{- 1}$ $=9.33ms^-1$

The velocity is the integral of the acceleration.

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

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

Plugging in the initial conditions

$v (0) = 8 = 0 - 0 + C$ $v(0)=8=0-0+C$

So,

$C = 8$ $C=8$

$v (t) = 2 t - \frac{1}{3} t^{3} + 8$ $v(t)=2t-1/3t^3+8$

We can calculate the mean speed

$(2 - 0) ¯ v = \int_{0}^{2} (2 t - \frac{1}{3} t^{3} + 8) d t$ $(2-0)barv=int_0^2(2t-1/3t^3+8)dt$

$= {[\frac{2}{2} t^{2} - \frac{1}{12} t^{4} + 8 t]}_{0}^{2}$ $=[2/2t^2-1/12t^4+8t]_0^2$

$= (4 - \frac{4}{3} + 16) - (0)$ $=(4-4/3+16)-(0)$

$= \frac{56}{3}$ $=56/3$

Therefore,

$¯ v = \frac{1}{2} \cdot \frac{56}{3} = \frac{28}{3} = 9.33 m s^{- 1}$ $barv=1/2*56/3=28/3=9.33ms^-1$

What is the average speed of an object that is moving at 8 m/s8ms8 m/s at t=0t=0t=0 and accelerates at a rate of a(t) =2-t^2a(t)=2−t2a(t) =2-t^2 on t in [0,2]t∈[0,2]t in [0,2]?