Given acceleration function a=ce^-t, find average velocity?

Question

An object starts from rest at t = 0 and accelerates so that a =ce^−t
where c is a positive
constant.
(a) Find the average velocity over the first t seconds.
(b) What happens to this average velocity as t becomes very large?

Answer 1

A) Velocity is the integral of acceleration.



V = -c*e^-t + constant. The constant is zero in this case because v = 0 at t = 0.



B) as t goes to infinity, the exponential term goes to 1 so V will go to -c.

Answer 2

a = ce^(-t)

v = -ce^(-t) + c (since v(0) = 0)



v = c(1 - e^(-t))



s = ce^(-t) + ct - c (assuming that s(0) = 0)

s = c(e^(-t) + t - 1)



s(0) = 0

s(t) = c(e^(-t) + t - 1)



a) average velocity = distance / time = c(e^(-t) + t - 1) / t



b) as t ==> infinity, average velocity goes to c

(e^(-t)) goes to 0 as t ==> inf

so you have e^(-t) / t ==> 0

t / t = 1

1 / t ==> 0



so the limit as t gets big is c(0 + 1 + 0) = c

Answer 3

Average velocity = integral from t = 0 to t = t of ce^-x dx, = c(1-e^-t)/t.

as t is large, average velcity becomes 0