What is the average speed of an object that is still at $t=0$ and accelerates at a rate of $a(t) = 3t^2-t+4$ from $t in [0,4]$ ?

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Answer 1 · 2017-07-10T09:18:38

The average speed is $=21.3ms^-1$

The speed is the integral of the acceleration.

$a(t)=3t^2-t+4$

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

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

Plugging in the initial conditions, $t=0$

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

Therefore,

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

The average speed is

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

$4barv=[1/4t^4-1/6t^3+2t^2]_0^4$

$=(64-32/3+32)-(0)$

$=256/3$

$barv=64/3=21.3ms^-1$

What is the average speed of an object that is still at t=0t=0 and accelerates at a rate of a(t) = 3t^2-t+4a(t) = 3t^2-t+4 from t in [0,4]t in [0,4]?