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!

Kinematics question, easier way to do this?

  1. May 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Two cars are travelling in a straight line, in the same direction, along a multi-lane highway in adjacent lanes. Car A starts 100 m behind Car B and both cars are initially travelling with a velocity of 20 m s−1 . Car B maintains a constant velocity but Car A begins to accelerate at 2.5 m s−2 . How long will it take for Car A to draw level with Car B (in s)?
    (A) 2.7 (B) 4.3 (C) 6.1 (D) 8.9 (E) 10
    2. Relevant equations
    vavg=d/t
    a=change in v/t
    vf=vi+at
    Vf^2=vi^2 + 2*a*d
    d=vi t + 0.5at^2
    d=(vf+vi)/2 * t

    3. The attempt at a solution
    I broke up the question into time intervals. I started with seeing how long it took car B to reach the initial 100m, then used that time to see how far car A gone in that time interval. I continued to break up the question like this and it became very tedious as I got closer to the 8.9s correct answer, and was wondering if there was a simpler way to approach this question? Thanks
     
  2. jcsd
  3. May 12, 2017 #2
    You should focus on the difference in speed between the two cars, that's how car A is going to close the gap
     
  4. May 12, 2017 #3

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Write two expressions, one for each car, giving the position at any time t. Subtract to get the distance between cars at any time t. Find at what specific time this distance becomes zero.
     
  5. May 12, 2017 #4

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What would happen if both cars started with a velocity of ##0m/s##?
     
  6. May 12, 2017 #5
    Wow, thank you, that worked!
    Just wondering how you went about solving it that way? Like, where does the cars starting at 0m/s come from? Is it because they started at the same velocity?
     
  7. May 12, 2017 #6

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, the constant velocity with which they start makes mo difference to the rate at which the second car catches the first.

    More formally, you can say you are using a reference frame moving along with the initial velocity.

    A useful trick!
     
  8. May 12, 2017 #7
    Consider it this way, as long as the cars have the same velocity, every change in the position of one of them will be matched by the other car, so you should just consider the difference in their speeds.

    If a car goes 10 m/s and the other goes at 11 m/s and they start in the same position it will take 10 seconds for the faster car to get 10 m ahead of the first one
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted