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

2 Answers
Apr 20, 2018

Answer:

The average speed is #=2ms^-1#

Explanation:

The speed is the integral of the acceleration

#a(t)=2t^2-2t+2#

#v(t)=int(2t^2-2t+2)dt#

#=2/3t^3-t^2+2t+C#

Plugging in the initial conditions

#v(0)=0#

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

#C=0#

The average speed on the interval #[0,2]# is

#(2-0)barv=int_0^2(2/3t^3-t^2+2t)dt#

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

#=(8/3-8/3+4)-(0)#

#2barv=4#

#barv=2ms^-1#

Apr 20, 2018

Given the acceleration of the object at t th instant as

#a(t) = 2t^2-2t+2#

#=>(dv(t))/(dt) = 2t^2-2t+2#

#=>v(t)= int(2t^2-2t+2)dt+c#, where c is integration constant.

#=>v(t)= 2/3t^3-2/2t^2+2t+c#

Given that object is still at #t=0#, we get #v(0)=0#

Hence #v(0)= 0=2/3xx0-2/2xx0+2xx0+c#

So #c=0#
Hence
#v(t)= 2/3t^3-t^2+2t#

#=>(ds(t))/(dt)= 2/3t^3-t^2+2t#

So displacement in the time interval #tin[0,2]#
#=>s(t)=int_0^2( 2/3t^3-t^2+2t)dt#

#=>s(t)=[ 2/12t^4-1/3t^3+2/2t^2]_0^2#

#=>s(t)=[ 2/12*2^4-1/3*2^3+2^2]#

#=>s(t)=[ 8/3-8/3+4]=4#unit

Hence average velocity #barv="displacement"/"time interval"=4/2=2# unit/s