What is the average speed of an object that is still at $t=0$ and accelerates at a rate of $a(t) = 12-3t$ from $t in [0, 4]$ ?

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Answer 1 · 2017-08-03T09:12:26

The average speed is $=16ms^-1$

The speed is the integral of the acceleration

$a(t)=12-3t$

$v(t)=int(12-3t)dt=12t-3/2t^2+C$

Plugging in the initial conditions, $v(0)=0$

Therefore,

$0=0-0+C$

So,

$v(t)=12t-3/2t^2$

The average speed is

$4barv=int_0^4(12t-3/2t^2)dt=[12/2t^2-1/2t^3]_0^4$

$=(96-32)-(0)=64$

$barv=64/4=16ms^-1$

What is the average speed of an object that is still at t=0t=0 and accelerates at a rate of a(t) = 12-3ta(t) = 12-3t from t in [0, 4]t in [0, 4]?