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

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Answer 1 · 2016-06-22T08:20:29

$v_a=5.67" "m/s$

$"we can solve either using the average value of a function,"$
$"or using definition of the average speed"$

$"solution using definition of average speed:"$
$"................................................................$

$v(t)=int a(t)*d t$

$v(t)=int (t^2-t)*d t$

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

$"So " t=0" ; "v=3 m/s" ; "C=3$

$v(t)=t^3/3-t^2/2+3" *"$

$Delta s=int _0^4 v(t)*d t$

$Delta s=int_0^4 (t^3/3-t^2/2+3)d t$

$Delta s=|1/3*1/4*t^4-1/2*1/3*t^3+3*t|_0^4$

$Delta s=(1/12*4^4-1/6*4^3+3*4)-(1/12*0-1/6*0+3*0)$

$Delta s=(256/12-64/6+12)-0$

$Delta s=(256-128+144)/12=272/12" m , total distance covered(t=0;t=4)"$

$v_a=("total distance covered")/("given time interval")" definition of average velocity"$

$v_a=(272/12)/(4-0)=272/(4*12)=272/48=5.67" "m/s$

$"solution using the average value theorem:"$
$"................................................................$

$"using speed v(t) * function above"$
$v_a=1/(4-0)int_0^4 v(t)*d t$

$v_a=1/4int_0^4 (t^3/3-t^2/2+3)*d t$

$v_a=1/4[|1/3*1/4*t^4-1/2*1/3*t^3-3*t|_0^4]$

$v_a=1/4[(1/12*4^4-1/6*4^3+3*4)-(0)]$

$v_a=1/4[272/12]$

$v_a=5.67" "m/s$

What is the average speed of an object that is moving at 3 m/s3 m/s at t=0t=0 and accelerates at a rate of a(t) =t^2-ta(t) =t^2-t on t in [0,4]t in [0,4]?