What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $1 \frac{m}{s}$ $1 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 2 t^{2} - 1$ $a(t) =2t^2-1$ on $t \in [0, 3]$ $t in [0,3]$ ?

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Answer 1 · 2018-01-16T07:07:15

The average speed is $= 4 m s^{- 1}$ $=4ms^-1$

The speed is the integral of the acceleration

$a (t) = 2 t^{2} - 1$ $a(t)=2t^2-1$

Then,

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

Plugging in the initial conditions at $t = 0$ $t=0$

$v (0) = 1 = 0 - 0 + C$ $v(0)=1=0-0+C$

$C = 1$ $C=1$

Therefore,

$v (t) = \frac{2}{3} t^{3} - t + 1$ $v(t)=2/3t^3-t+1$ in the interval $[0, 3]$ $[0,3]$

After that $t > 3$ $t>3$ , assume that the speed is constant

$v (3) = \frac{2}{3} \cdot 3^{3} - 3 + 1 = 4 m s^{- 1}$ $v(3)=2/3*3^3-3+1=4ms^-1$

The average speed on the interval $[0, 5]$ $[0,5]$ is

$5 ¯ v = \int_{0}^{3} (\frac{2}{3} t^{3} - t + 1) d t + (4 \cdot 2)$ $5barv=int_0^3(2/3t^3-t+1)dt+(4*2)$

$= {[\frac{1}{6} t^{4} - \frac{1}{2} t^{2} + t]}_{0}^{4} + 8$ $=[1/6t^4-1/2t^2+t]_0^3+8$

$= (\frac{1}{6} \cdot 3^{4} - \frac{1}{2} \cdot 3^{2} + 3) + 8$ $=(1/6*3^4-1/2*3^2+3)+8$

$= 20$ $=20$

The average speed is

$¯ v = \frac{20}{5} = 4 m s^{- 1}$ $barv=20/5=4ms^-1$

What is the average speed, on t in [0,5]t∈[0,5]t in [0,5], of an object that is moving at 1 m/s1ms1 m/s at t=0t=0t=0 and accelerates at a rate of a(t) =2t^2-1a(t)=2t2−1a(t) =2t^2-1 on t in [0,3]t∈[0,3]t in [0,3]?