What is the average speed, on $t in [0,5]$ , of an object that is moving at $5 m/s$ at $t=0$ and accelerates at a rate of $a(t) =t^2-2t$ on $t in [0,4]$ ?

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Answer 1 · 2017-11-16T07:06:13

The average speed is $=6.07ms^-1$

The speed is the integral of the acceleration

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

Then,

$v(t)=int(t^2-2t)dt=t^3/3-2t^2/2+C$

Plugging in the initial conditions at $t=0$

$v(0)=5=0-0+C$

$C=5$

Therefore,

$v(t)=t^3/3-t^2+5$ in the interval $[0,4]$

After that $t>4$ , assume that the speed is constant

$v(4)=4^3/3-4^2+5=64/3-16+5=10.33ms^-1$

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

$5barv=int_0^4(t^3/3-t^2+5)dt+10.33*1$

$=[t^4/12-t^3/3+5t]_0^4+10.33$

$=(4^2/12-4^3/3+5*4)+10.33$

$=30.33$

The average speed is

$barv=30.33/5=6.07ms^-1$

What is the average speed, on t in [0,5]t in [0,5], of an object that is moving at 5 m/s5 m/s at t=0t=0 and accelerates at a rate of a(t) =t^2-2ta(t) =t^2-2t on t in [0,4]t in [0,4]?