1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How long until she passes you?

  1. Sep 25, 2016 #1
    1. The problem statement, all variables and given/known data
    You leave a spotlight with an acceleration of 4.9m/s2[fwd] until reaching a speed of 49m/s. A police officer took off from the same intersection 2sec after. She accelerates at 9.8m/s2. How long after this does she pass you? How fast is she going at this time?


    2. Relevant equations


    3. The attempt at a solution
    4.9t2 = 9.8 x (t-2)2
     
  2. jcsd
  3. Sep 25, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    You'll need to show more of your attempt and describe what you are trying to do there. What does your equation represent?

    What type of equations are relevant to the problem? Can you list them?
     
  4. Sep 28, 2016 #3
    Silly question: Does this require calculus and did he write his equations in calculus?

    I haven't done calculus yet..
     
  5. Sep 28, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    No calculus required. Just basic kinematic equations. The OP might have made things more clear if he'd availed himself of the x2 icon in the edit panel to create superscripts. It's can be hard to tell sometimes if a '2' is meant to be a constant multiplying something or an exponent. :smile:
     
  6. Sep 28, 2016 #5
    Okay, let me attempt it:

    at a = 4.9 m/s /s there is a constant acceleration in the + direction until v = 49 m/s
    a = m/s /s & v=m/s
    a=Δv/t
    Thus cancel to get time =
    10 s

    r = ut + 1/2a t^2
    r = 1/2 a t^2 = 245m

    ( i completely forgot about taking off 2s after)

    at t = 2 we can find velocity and distance, but from here i do have a clue how we can find t for when r1 = r2
    (if this is how i should do it)
     
    Last edited: Sep 28, 2016
  7. Sep 28, 2016 #6

    gneill

    User Avatar

    Staff: Mentor

    In your work you didn't account for the 2 second delay before the cop started to accelerate. Also, if it does take the cop 50 seconds to reach the car, the car will have long ago entered its constant cruise speed phase of its motion. That would complicate things...

    1. Not really. Unless you're not interested in great accuracy, then you might do it graphically. The area under a velocity versus time graph is distance.
    2. Given that v = at, just integrate with respect to t. The integration constant is the initial velocity.
     
  8. Sep 28, 2016 #7
    This would be correct if the police officer caught up with you in under 10 sec. Check this possibility by solving the equation for t to see if that is actually the case.
     
  9. Sep 28, 2016 #8
    I completely forgot, mb. Also, I edited the question instead of replying. Exausted brain doesn't work very well...
    Anyway, my edit was:


    Okay, let me attempt it:

    at a = 4.9 m/s /s there is a constant acceleration in the + direction until v = 49 m/s
    a = m/s /s & v=m/s
    a=Δv/t
    Thus cancel to get time =
    10 s

    r = ut + 1/2a t^2
    r = 1/2 a t^2 = 245m

    ( i completely forgot about taking off 2s after)

    at t = 2 we can find velocity and distance, but from here i do have a clue how we can find t for when r1 = r2
    (if this is how i should do it)
     
  10. Sep 28, 2016 #9
    Shouldn't we give the OP a chance to respond?
     
  11. Sep 28, 2016 #10
    I want to play too (:
     
  12. Sep 28, 2016 #11

    gneill

    User Avatar

    Staff: Mentor

    You're probably right. But then again, the OP made the one post to start this thread three days ago and hasn't resurfaced.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted