What is the average speed, on $t in [0,5]$ , of an object that is moving at $8 m/s$ at $t=0$ and accelerates at a rate of $a(t) =5t^2-3t$ on $t in [0,4]$ ?

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Answer 1 · 2017-11-29T07:45:06

The average speed is $=38.67ms^-1$

The speed is the integral of the acceleration

$a(t)=5t^2-3t$

Then,

$v(t)=int(5t^2-3t)dt=5/3t^3-3/2t^2+C$

Plugging in the initial conditions at $t=0$

$v(0)=8=0-0+C$

$C=8$

Therefore,

$v(t)=5/3t^3-3/2t^2+8$ in the interval $[0,4]$

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

$v(4)=5/3*4^3-3/2*4^2+8=320/3-28+8=86.67ms^-1$

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

$5barv=int_0^4(5/3t^3-3/2t^2+8)dt+86.67*1$

$=[5/12t^4-1/2t^3+8t]_0^4+86.67$

$=(5/12*4^2-4^3/2+8*4)+86.67$

$=193.3$

The average speed is

$barv=193.3/5=38.67ms^-1$

What is the average speed, on t in [0,5]t in [0,5], of an object that is moving at 8 m/s8 m/s at t=0t=0 and accelerates at a rate of a(t) =5t^2-3ta(t) =5t^2-3t on t in [0,4]t in [0,4]?