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Help With Problems

  1. Jun 19, 2008 #1
    After refueling, car has an acceleration whose magnitude is 6.8 m/s^2, after 4 sec. he enters the speedway. At the same instant, another car and traveling at a constant speed of 62 m/s overtakes and passes the entering car. If the entering car maintains its acceleration, how much time is required for it to catch the other car?
  2. jcsd
  3. Jun 19, 2008 #2


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    You will want to set up separate position equations for each car. Call the point at the end of the runway from the maintenance pit, where the entering car returns to the race track, x = 0 ; also call the time when the entering car returns, and the moving car is just passing it, t = 0.

    At that moment, the car in the race is moving at a constant 62 m/sec. What would its position kinematic equation be (make it, say, x_B = ... ) ? The entering car has been accelerating at 6.8 m/(sec^2) for 4 seconds by t = 0 when it reaches x = 0. What would this car's position equation be ( x_A = ... ) ?

    When the entering car catches up to the car already in the race, you will have x_A = x _B . On setting your two equations equal, you will have something you can rearranging into a single quadratic equation you can solve for t . There will probably be two solutions, but since the equation only applies for t = 0 and afterwards, you can throw away any negative answer for t.
  4. Jun 21, 2008 #3
    Car 1's velocity upon entering the track:
    [tex]\ a = \frac{v - v_0}{t}[/tex]
    [tex]\ v = a·t + v_0 = (6.8\:m/s^{2})(4\:s) + 0 = 27.2\:m/s[/tex]

    Car 1's position on track at time t:
    [tex]\ d = vt + \frac{1}{2}at^{2} [/tex]

    Car 2's position on track at time t:
    [tex]\ d = vt [/tex]

    Set these positions equal to one another and solve for t:
    [tex]\ v(car 1)t + \frac{1}{2}at^{2} = v(car 2)t[/tex]
    [tex]\ v(car 1)t + \frac{1}{2}at^{2} - v(car 2)t = 0 [/tex]
    [tex]\ t[v(car 1) - v(car2) + \frac{1}{2}at] = 0 [/tex]

    t = 0 is the first solution that jumps out at you, but it's uninteresting, because it just says the car 2 overtook car 1 at the beginning, which we already know from the problem statement. So you want to t for the bracketed part:
    [tex]\ v(car 1) - v(car2) + \frac{1}{2}at = 0 [/tex]
    [tex]\ \frac{1}{2}at = v(car2) - v(car1) [/tex]
    [tex]\ t = 2\frac{v(car2) - v(car1)}{a} = 2\frac{62\:m/s - 27.2\:m/s}{6.8\:m/s^{2}} = 10.2\:s [/tex]
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