How do I antidifferentiate from acceleration to velocity?

a body has an initial displacement of 5 m and velocity 2m/s. find the displacement and velocity after 4 seconds given that #a=6/(t+1)^3#

(a=acceleration.)

i got v=1.88m/s and s = 5.6 m - but the answer is wrong apparently.

1 Answer
Øko
Jan 26, 2018

#v(4)~~2.27m/s#

#s(4)~~13.76m#

Explanation:

We have the relations

#s''(t)=v'(t)=a(t)#

Or for the use of this exercise

#inta(t)dt=v(t)# and #intv(t)dt=s(t)#

We know

#a(t)=6/(t+3)^3#, #v(0)=2#, #s(0)=5#

Find the velocity function

#v(t)=int6/(t+3)^3dt#

#=6*(-1/2)*1/(t+3)^2+C#

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

Evaluate the constant of integration by using #v(0)=2#

#2=-3/(0+3)^2+C<=>C=7/3#

Then we find #v(4)#

#v(4)=-3/(4+3)^2+7/3~~2.27#

Then we find the distance function

#s(t)=int-3/(t+3)^2+7/3dt#

#=3/(t+3)+7/3t+C#

Evaluate the constant of integration by using #s(0)=5#

#5=3/(0+3)+7/3*0+C<=>C=4#

Then we find #s(4)#

#s(4)=3/(4+3)+7/3*4+4~~13.76#

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