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

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Answer 1 · 2018-04-03T06:29:29

The average speed is $=13.25ms^-1$

The speed is the integral of the acceleration.

$a(t)=t^2-2t+4$

$v(t)=int(t^2-2t+4)dt=1/3t^3-t^2+4t+C$

Plugging in the initial conditions

$v(0)=8$

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

Therefore,

$v(t)=1/3t^3-t^2+4t+8$

The average speed is

$barv=1/(3-0)int_0^3v(t)dt=1/(3-0)int_0^3(1/3t^3-t^2+4t+8)dt$

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

$=1/3(81/12-9+18+24)$

$=13.25ms^-1$

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