- #1

customwelds

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## Homework Statement

A particle moving along a straight line decelerates according to a= -kv, where k is a constant and v is velocity. If it’s initial velocity at time t=0 is v0=4 m/s and its velocity at time t=2 s is v = 1 m/s, determine the time T and corresponding distance D for the particle speed to be reduced to one-tenth of its initial value

Answer : T = 3.32 s, D = 5.19 M

## Homework Equations

I think I figured out the value of the constant (k). However, I have no idea where to go from there. Below is my work on how I got to that point. I have tried integrating the new acceleration equation but have not had success further to find answers that are correct. Its very possible I'm not even on the right track so if anyone could please point me in the right direction I would be grateful.

## The Attempt at a Solution

equation: a=-kv ;; a=-k(dv/dt) ;; 1/v(dv) = -kdt ;; ln|v| = -kt +c

With that, solving for C we can use: ln|4| = -kt +c @ t=0 ;; ln|4| = 1.3863 = C

Then solve for k using: ln|1| = -kt + 1.3863 ;; k=0.69315

So from here the acceleration at t=0 and t=2 can be solved by using the velocities

a (@ t=0; v=4) = -.69315(4) = -2.7726m/s^2

a (@ t=2; v=1) = -.69315(1) = -.69315m/s^2

No idea where to go from here. :(