Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Determine the times at which the velocity=0

  1. Jan 3, 2006 #1
    the acce of a particle is a=A - 6t^2 where A is a constant. At t=0,the particle starts at x=8m with v=0. Knowing that at t=1s, v=30m/s.
    Determine the times at which the velocity=0

    pls help...
    the answer is 0 and 4s
     
  2. jcsd
  3. Jan 3, 2006 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    Now that you are given the acceleration of the particle, do you know how to find the velocity from the initial condition? (Remember: acceleration is the time derivative of the velocity).
     
  4. Jan 4, 2006 #3
    ya,i tried to get dv/dt=A-6t^2 to v-v(initial)=(At-2t^3) - (At[initial] - 2t^3[initial]).but i think it is wrong....
    what eqn do u get to form??
     
  5. Jan 4, 2006 #4
    [tex]v = At - 2t^{3} + C[/tex]

    You know v(0)=0 and v(1) = 30, so you can work out A and C. Once you have that, solve v = 0.
     
  6. Jan 4, 2006 #5
    for part b) what is the total distance travelled by the particle when t=5s??

    i try to integrate and get x=16t^2 - 1/2t^4 + 8 but when i try to do,the answer is wrong.

    the answer is 168.5...pls help
     
  7. Jan 4, 2006 #6

    Fermat

    User Avatar
    Homework Helper

    The particle first moves away from the origin until v = 0.
    How far has it travelled ? How long did it take ?

    How much further does it travel, on the way back, until t = 5 sec ?
     
  8. Jan 4, 2006 #7
    firstly is my eqn correct above??
    i try to find all the way from t=1 to 5 and found it goes back...what should i do
     
  9. Jan 4, 2006 #8

    Fermat

    User Avatar
    Homework Helper

    Yes, your eqn is correct.
    You just have two distances, the distance out and then a bit more, coming back, to find.

    Think of it like a particle in shm. It starts at the origin. Moves out a distance and then moves back towards the origin again.
    That is what this particle is doing, although the movement is certainly not shm.

    Find out when the velocity = 0. That is when the particle is at its max. distance from the origin. What is the value of t ?
    Use your eqn to find out the distance, x, from the origin. That is how far it has travelled so far.
    Now find its distance from the origin at t = 5. What distance has the particle moved back by ?
    Now what is the total distance moved ?
     
  10. Jan 4, 2006 #9
    i thought the problem stated that at t=0,x=8 and v=0 also.so it started at 8 for t=0.
    then at t=5s, i got 95.5

    is it correct??
     
  11. Jan 4, 2006 #10

    mukundpa

    User Avatar
    Homework Helper

    x in the equation is displacement, not the distance travelled. If you move 100 m and come back adistance 20 m, the total distance travelled is 120 m but the displacement is only 80 m. In the same way the particle is moving forward till v=0 and then comes back, you have to find distance travelled
     
  12. Jan 4, 2006 #11

    Fermat

    User Avatar
    Homework Helper

    Everything in your last post is correct, but ...

    The particle starts from the origin, moves out by about 130 odd metres, stops momentarily (v=0, t < 5) then moves back again towards the origin such that at t = 5, its distance from the origin is then 95.5 m
    You have to (first of all) find out its distance from the origin, when v=0.

    What's the eqn for velocity ?
     
  13. Jan 6, 2006 #12
    now when v=0,t=- or + 4s.so i substituted into the eqn x=16t^2 - 1/2t^4 + 8 , i got 120.
    then i try to take 120-95.5=24.5 and then plus 120 =144.5.However,the answer is incorrect.Where did i go wrong??
     
  14. Jan 6, 2006 #13

    Fermat

    User Avatar
    Homework Helper

    You got 120 by subtracting 8 instead of adding it!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook