What is the average speed of an object that is moving at $6 m/s$ at $t=0$ and accelerates at a rate of $a(t) =t^2-5t+3$ on $t in [0,3]$ ?

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Answer 1 · 2017-07-02T06:56:09

The average speed is $=5.25ms^-1$

The speed is the integral of the acceleration

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

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

$=1/3t^3-5/2t^2+3t+C$

Plugging in the initial conditions,

$v(0)=6$

$v(0)=0-0+0+C=6$

$=>$ , $C=6$

Therefore,

$v(t)=1/3t^3-5/2t^2+3t+6$

Let the average speed be $barv$

Then,

$(3-0)barv=int_0^3(1/3t^3-5/2t^2+3t+6)dt$

$=[1/12t^4-5/6t^3+3/2t^2+6t]_0^3$

$=(1/12*3^4-5/6*3^3+3/2*3^2+6*3)-(0)$

$=15.75$

$barv=15.75/3=5.25ms^-1$

What is the average speed of an object that is moving at 6 m/s6 m/s at t=0t=0 and accelerates at a rate of a(t) =t^2-5t+3a(t) =t^2-5t+3 on t in [0,3]t in [0,3]?