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Two cars at same speed, one passes the other

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data
    This question has gotten me stuck. I translated it from french to english so pardon that... I just get lost when you include two objects in physics.

    A car and a truck move initially in the same direction at 20 m / s, the truck with 38m ahead. The car accelerates at a constant rate of 2m / s ^ 2 exceeds the truck, and falls back into the right lane when it is 11m in front of the truck.
    a) How far has the truck traveled in that time?
    b) For what value of constant acceleration of the truck it and the car will they have a velocity that differs by only 5 m / s when they intersect? It is assumed that the truck begins to accelerate when a lead of 38m on the car.


    2. Relevant equations
    d= Vi*t + (0,5)at^2
    d= distance of the car


    3. The attempt at a solution
    I tried
    But I keep getting negative answers! Someone give me a boost or help me start it! Please!

    ...
    answers:
    a) 140m
    b) a(car) > 1,67 > s^2
     
  2. jcsd
  3. Oct 17, 2011 #2
    I get the same answer for part a. For part b you know that when they intersect the position (x value) for both the car and truck are equal (xcar=xtruck) where you can apply this equation for both the car and truck

    [tex]
    x = x_0 +v_0t+\frac{1}{2}at^2
    [/tex]

    You also know that at this point (same t) the velocities differ by 5m/s (vcar-vtruck = 5m/s) where you can apply this equation for both the car and truck

    [tex]
    v = v_0 +at
    [/tex]

    This should give you two equations and two unknowns (t and a).
     
  4. Oct 17, 2011 #3
    Thanks! But, Im still stuck on a) though... I cant see how i can mesure the distance persued bby the truck..
     
  5. Oct 17, 2011 #4
    I solved for the time it took for the car to get 11m in front of the truck and then put that into the equation for the truck (not accelerating xtruck = xo + vot). I also got 140m, so unless I'm wrong you have the right answer.
     
  6. Oct 18, 2011 #5
    I didnt find those answers, those are the correct answers provided by my book.
     
  7. Oct 18, 2011 #6
    I just plugged in these in the equation for a) and found the time to catch up with the truck, i think...:
    d = Vi*t + (0,5)at^2
    38 = 20t + (0,5)2t^2

    but i dont think thats right, because the truck is always moving forward, not staying at 38m...
     
  8. Nov 8, 2011 #7
    I got some help from a tutor and Seems ive been a tad blind, thank you for helping, means a lot!
     
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