What is the average speed of an object that is moving at $5 \frac{m}{s}$ $5 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 20 - t^{2}$ $a(t) =20-t^2$ on $t \in [1, 5]$ $t in [1,5]$ ?

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Answer 1 · 2017-04-04T10:30:40

The average speed is $= 52 m s^{- 1}$ $=52ms^-1$

The velocity is the integral of the acceleration.

$a (t) = 20 - t^{2}$ $a(t)=20-t^2$

$v (t) = \int (20 - t^{2}) d t$ $v(t)=int(20-t^2)dt$

$= 20 t - \frac{t^{3}}{3} + C$ $=20t-t^3/3+C$

We apply the initial conditions to find $C$ $C$

$v (0) = 5$ $v(0)=5$

Therefore, $C = 5$ $C=5$

So,

$v (t) = 20 t - \frac{t^{3}}{3} + 5$ $v(t)=20t-t^3/3+5$

We can calculate the average value of the velocity

$(5 - 1) ¯ v = \int_{1}^{5} (20 t - \frac{t^{3}}{3} + 5) d t$ $(5-1)barv=int_1^5(20t-t^3/3+5)dt$

$= {[\frac{20}{2} t^{2} - \frac{t^{4}}{12} + 5 t]}_{1}^{2}$ $=[20/2t^2-t^4/12+5t]_1^5$

$= (250 - \frac{625}{12} + 25) - (10 - \frac{1}{12} + 5)$ $=(250-625/12+25)-(10-1/12+5)$

$= (\frac{2675}{12}) - (\frac{179}{12})$ $=(2675/12)-(179/12)$

$= 208$ $=208$

So,

$¯ v = \frac{208}{4} = 52 m s^{- 1}$ $barv=208/4=52ms^-1$

What is the average speed of an object that is moving at 5 m/s5ms5 m/s at t=0t=0t=0 and accelerates at a rate of a(t) =20-t^2a(t)=20−t2a(t) =20-t^2 on t in [1,5]t∈[1,5]t in [1,5]?