What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $10 \frac{m}{s}$ $10 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 2 t^{2} - 10$ $a(t) =2t^2-10$ on $t \in [0, 3]$ $t in [0,3]$ ?

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Answer 1 · 2017-08-06T06:46:00

The average speed is $= 1.6 m s^{- 1}$ $=1.6ms^-1$

The speed is the integral of the acceleration

$a (t) = 2 t^{2} - 10$ $a(t)=2t^2-10$

$v (t) = \frac{2}{3} t^{3} - 10 t + C$ $v(t)=2/3t^3-10t+C$

Plugging in the initial conditions

$v (0) = 0 - 0 + C = 10$ $v(0)=0-0+C=10$

Therefore,

$v (t) = \frac{2}{3} t^{3} - 10 t + 10$ $v(t)=2/3t^3-10t+10$ on the interval $I = [0, 3]$ $I=[0,3]$

The total distance travelled in $5 s$ $5s$ is

$d = \int_{0}^{3} (\frac{2}{3} t^{3} - 10 t + 10) d t + v (3) \cdot 2$ $d=int_0^3(2/3t^3-10t+10)dt+v(3)*2$

$= {[\frac{1}{3} t^{4} - 5 t^{2} + 10 t]}_{0}^{4} + (18 - 30 + 10) \cdot 2$ $=[1/3t^4-5t^2+10t]_0^3+(18-30+10)*2$

$= (27 - 45 + 30) + (- 4)$ $=(27-45+30)+(-4)$

$= 8$ $=8$

The average speed is

$¯ v = \frac{8}{5} = 1.6 m s^{1}$ $barv=8/5=1.6ms^1$

What is the average speed, on t∈[0,5]t in [0,5], of an object that is moving at 10ms10 m/s at t=0t=0 and accelerates at a rate of a(t)=2t2−10a(t) =2t^2-10 on t∈[0,3]t in [0,3]?