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: Very simple kinematics question

  1. Aug 28, 2010 #1
    1. The problem statement, all variables and given/known data
    i haven't taken physics since freshman year of high school and this is chapter 1 in physics. I forgot i am sorry. I know it's really easy

    A sports car moving at constant speed travels 120m in 5.3s , then brakes and comes to a stop in 3.8s .
    What is the magnitude of its acceleration in m/s2

    2. Relevant equations

    a= ∆v/∆t
    v=∆x/∆t
    or a=dv/dt

    i have no clue which equations to use. hint please.


    3. The attempt at a solution
    v= ∆x/∆t
    x1= 120(5.3)= 636
    x0= 0
    ∆t= 3.8-5.3
    average V= 162.1 m/s2

    i have to get the magnitude of it's acceleration; is that the same as it's average acceleration? i don't know what i am doing. sorry
     
    Last edited: Aug 28, 2010
  2. jcsd
  3. Aug 28, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You'll need both of these equations.

    He's only accelerating for 3.8 seconds. (For the 5.3 seconds before that, he was moving at constant speed.)

    What's the initial speed before he begins accelerating? (Figure it out from the given data.)
    The final speed?

    The change in speed? The time it takes?

    Then you can calculate the acceleration.
     
  4. Aug 28, 2010 #3
    thank you! i had a huge brain fart
     
  5. Aug 28, 2010 #4
    but i might be having one again, so just to make sure:


    v=∆x/∆t
    x2=120/5.3 (3.8) = 86.0m; x1= 120m so 120+86= 206 therefore ∆x= 86-206 = -120
    ∆t=3.8- 5.3= 1.5

    v=80m/s^2

    a=80/1.5
    a=53.3??

    is this right?
     
  6. Aug 28, 2010 #5

    Doc Al

    User Avatar

    Staff: Mentor

    No, not right. Please answer each of the questions I asked in my last post. That will lead you step by step to the answer.
     
  7. Aug 28, 2010 #6
    the initial speed before he accelerated is 22.6m/s (120/5.3)
    the final speed is 0 because he came to a stop
    the change in speed is -22.6
    the time for both would be -1.5 (3.8-5.3)
    so a=∆v/∆t
    -22.6/-1.5 = 15.0?
     
  8. Aug 28, 2010 #7

    Doc Al

    User Avatar

    Staff: Mentor

    Perfect.
    Careful: The time you need here is just the time for the acceleration. That's given--no need to calculate anything. (How long does it take to come to a stop?)

    Fix that and you've got it.
     
  9. Aug 28, 2010 #8
    so it is -22.6/3.8 which is 5.9 and the negative is to just denote a deceleration right? should it be written as a negative or is 5.9 ok.
     
  10. Aug 28, 2010 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Right. The negative sign just means that the acceleration is opposite to the velocity.
    You only want the magnitude, which is always positive.
     
  11. Aug 28, 2010 #10
    thank you very much, i feel like i need to pay you for your help. Thank you thank you. You are awesome.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook