Dynamics HW problem giving me a real headache

[customwelds]

Hi everyone at PF. Thanks for taking time to read my HW problem. It's really neat that you are willing to take time from your busy schedules to help out students such as myself. I really apprecciate it!

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. :(

 

[kuruman]

The Attempt at a Solution

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

Does this last equation make sense? You are doing physics not math, so when you integrate, there need to be limits of integration on both sides. What are those limits of integration? Furthermore, in physics, arguments of logarithms, sines, cosines, etc. have to be dimensionless. Yours isn't.

 

[collinsmark]

Does this last equation make sense? You are doing physics not math, so when you integrate, there need to be limits of integration on both sides. What are those limits of integration? Furthermore, in physics, arguments of logarithms, sines, cosines, etc. have to be dimensionless. Yours isn't.

Actually, in my opinion it's a matter of preference. You could evaluate the definite integral using initial conditions as limits, or evaluate the indefinite integral and later solve for the arbitrary constant using initial conditions (the latter is what the OP did). Both are fine approaches.

[Another edit: Oh, and regarding the units (the OP's arguments weren't dimensionless). It can be shown that the units sort of pop out with the arbitrary constant. Using different units for v merely cause the arbitrary constant to reflect the difference. It's a property of log(ab) = loga + logb.]

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

Okay, so far so good.

Eventually, you'll probably want to do some algebra and solve for v as a function of t. But you can solve for that later if you wish.

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. :(

Wait, what? Acceleration?

You don't need to find acceleration in order to solve this problem.

You already have a full relationship between velocity and time. And you already know what the initial v₀ is (4 m/s). So just plug in 0.1v₀ in for v, and solve for t!

For the next step you'll need to integrate to find distance, x(t) (complete with some algebra too). And you'll end up with another arbitrary constant if you use the indefinite integral. It's assumed in the problem statement that it's looking for the change in distance starting from t = 0. So that implies that x = 0 at t = 0.

[Edit: customwelds, I just noticed that in one of your listed equations, you had "a = -k(dv/dt)." I think you meant, dv/dt = -kv. I assume that was just a typo/simple mistake, because later your relationship turns out to be correct. And also, welcome to Physics Forums!]