- #1
customwelds
- 1
- 0
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!
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
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.
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. :(
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. :(