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]#?

1 Answer
Jun 22, 2016

Answer:

#v_a=5.67" "m/s#

Explanation:

#"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#