What is the average speed of an object that is not moving at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 8 - t$ $a(t) =8-t$ on $t \in [2, 5]$ $t in [2,5]$ ?

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Answer 1 · 2017-11-25T07:12:55

The average speed is $= 21.5 m s^{- 1}$ $=21.5ms^-1$

The speed is the integral of the acceleration

$a (t) = 8 - t$ $a(t)=8-t$

$v (t) = \int (8 - t) d t = 8 t - \frac{1}{2} t^{2} + C$ $v(t)=int(8-t)dt=8t-1/2t^2+C$

Plugging in the initial conditions

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

so,

$v (0) = 8 \cdot 0 - \frac{1}{2} \cdot O + C = 0$ $v(0)=8*0-1/2*O+C=0$

$C = 0$ $C=0$

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

The average speed on the interval $t \in [2, 5]$ $t in [2,5]$ is

$(5 - 2) ¯ v = \int_{2}^{5} (8 t - \frac{1}{2} t^{2}) d t$ $(5-2)barv=int_2^5(8t-1/2t^2)dt$

$3 ¯ v = {[\frac{8}{2} t^{2} - \frac{1}{6} t^{3}]}_{2}^{2}$ $3barv=[8/2t^2-1/6t^3]_2^5$

$3 ¯ v = (4 \cdot 25 - \frac{1}{6} \cdot 125) - (4 \cdot 4 - \frac{1}{6} \cdot 8)$ $3barv=(4*25-1/6*125)-(4*4-1/6*8)$

$3 ¯ v = 84 - \frac{125}{6} + \frac{4}{3} = 64.5$ $3barv=84-125/6+4/3=64.5$

$¯ v = \frac{64.5}{3} = 21.5 m s^{- 1}$ $barv=64.5/3=21.5ms^-1$

What is the average speed of an object that is not moving at t=0t=0 and accelerates at a rate of a(t)=8−ta(t) =8-t on t∈[2,5]t in [2,5]?