The position of a particle moving along the x-axis can be expressed x(t) = v0/k * ( 1-e^(-kt) )?

Question

x(t) = v0/k * ( 1 - e^(-kt) ), where v0 and k are constants.

a) If k = 2, how long does it take the velocity to slow to 1/100 of its original velocity?

b) Show that the acceleration is directed opposite to the velocity and has a magnitude proportional to the speed.

Thanks in advance.

Thank you so much everyone. I had a feeling it required a derivative to get the velocity. And for Daniel, my teacher didn't assign this problem, I'm just looking ahead at more advanced problems to get an idea of what's in store for me. All my teacher covered so far is projectile motion.

Answer 1

velocity = dx/dt = v0/k*ke^(-kt) = v0 e^(-kt)

At t= 0, v = v0 so we want to find t such that v = v0/100.

v0/100 = v0 e^(-kt) = 1/100 take ln of both sides

-ln(100) = -kt ---> t = ln(100)/k = 2.3 seconds if k = 2

Acceleration = dv/dt = -v0*k*e^(-kt) = -kv since v=v0*e^(-kt)

Hence a = -kv so acceleration is opposite (minus sign) velocity and proportional (k = constant of proportionality) to velocity.

Answer 2

x(t)= Vo/k (1-e^-2t) = Vo/2 (1-e^-2t)
x´(t)=v(t) =Vo/2*( 2e^-2t)= Vo e^-2t
Voe^-2t= 1/100* Vo so
e^-2t= 1/100 and -2t = ln (1/100)
t= 2.30 ( in the units of time)
v´(t) = a(t) = -2Vo e^-2t which has opposite sign to v(t) and is proportional to I Vo e^-2t I which is the speed

Answer 3

I don't see that you've made the slightest effort to work this out yourself.

"Gee, I'd love to help you, but I really don't want to"
Phoebe from "Friends"