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)^3a=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/sv(4)≈2.27ms

s(4)~~13.76ms(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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