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!

Braking speed and distance traveled

  1. Sep 18, 2016 #1
    1. The problem statement, all variables and given/known data
    The acceleration of a particular car during braking has magnitude bt, where t is the time in seconds from the instant the car begins braking, and b = 2.2 m/s3 .If the car has an initial speed of 60 m/s , how far does it travel before it stops?


    2. Relevant equations:
    I know that basic kinematic equations will not work for this equation because acceleration is not constant. Therefore the equations have to be integrated from the given equation of acceleration


    3. The attempt at a solution:
    when i integrated i got the equations:
    a=2.2t

    v= 2.2/2 t^2 +60

    x= 2.2/6 t^3 +60t

    I'm trying to figure out how to isolate for t in order to solve for distance, but when i set the equation for distance equal to zero i ended up getting the wrong answer while solving for t. if someone can point me in the right direction on how to solve for t, i know how to finish the problem solving for distance. i just am unsure about how to get time
     
  2. jcsd
  3. Sep 18, 2016 #2

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi Emily081715 and welcome to PF

    :welcome:

    Your expression for v says that v increases with time. Is that what happens to a car that is braking?
    Also, x is not zero when the car stops. What is zero when that happens?
     
  4. Sep 18, 2016 #3
    should my equations say -60? If that was the case shouldn't my acceleration be negative as well since its braking and slowing down?
    Also, if x isn't zero when the car stops, is the final velocity zero ? that would mean i need to set that equation equal to zero to solve for time
     
  5. Sep 18, 2016 #4

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    When t is equal to zero, v = 60 m/s as it should. When t is greater that zero, v starts to decrease. Where should the negative sign go?
     
  6. Sep 18, 2016 #5
    the negative sign goes infront of the 60. making the equation v= 2.2/2t^2- 60t
     
  7. Sep 18, 2016 #6

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    But now when you set t = 0 in your expression, you get zero, not 60 m/s.
     
  8. Sep 18, 2016 #7
    should it then be infront of 2.2
     
  9. Sep 18, 2016 #8

    David Lewis

    User Avatar
    Gold Member

    The magnitude of acceleration is positive even when the car is slowing down. The plus and minus sign give you the directional sense of the quantity. If we say the sense of the car's velocity is positive then the sense of the acceleration would be negative.
     
  10. Sep 18, 2016 #9

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. Leave the 60 as you had it. Now can you find the time of travel? It is the same as the time it takes the car to stop.
     
  11. Sep 18, 2016 #10

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's another of saying that (in 1-dimension), when the velocity and the acceleration have the same sign, the speed is increasing; when they have opposite signs, the speed is decreasing.
     
  12. Sep 18, 2016 #11
    Okay so i changed the sign of acceleration and used the equation v=-2.2/2 t^2 +60 to solve for t. i got (10 √66)/11 for time. when plugging that answer in for distance i got 295.4 meters. does this answer seem correct?
     
  13. Sep 18, 2016 #12

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It is correct.
     
  14. Sep 18, 2016 #13
    thanks for the help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Braking speed and distance traveled
Loading...