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(Acceleration) Car and Truck Problem

  1. Aug 12, 2011 #1
    Here's the problem:

    1. An automobile and a truck start from rest and at the same instant, with the automobile initially at some distance behind the truck. The truck has a constant acceleration of 2.10 m/s2, and the automobile an acceleration of 3.4 m/s2. The automobile overtakes the truck after the truck has moved 50.0 m. (a) How much time does it take the automobile to overtake the truck? (b) How far was the automobile behind the truck initially? (c) Draw the figure of the problem



    2. Equations and Data : a=[itex]\frac{dV}{dt}[/itex] ; V = [itex]\frac{dx}{dt}[/itex]

    CAR:
    Acceleration = 3.4m/s2

    TRUCK:
    Acceleration = 2.10m/s2
    Distance Traveled at time of overtaking = 50m



    3. The attempt at a solution:
    For a.)

    Reading the problem, The time that the car overtook the truck should be the same as the overall time of the truck to cover 50m. So finding for t of the truck, I would have

    [itex]\int[/itex]2.10dt = [itex]\int[/itex]dV

    --Integrating this i would get

    2.10t = V

    I can solve for time using the equation for Velocity and 50 m as the displacement

    V = [itex]\frac{dx}{dt}[/itex]

    [itex]\int[/itex]2.10t dt=[itex]\int[/itex] dx

    --integrating this this

    1.05t2 = x

    -- substituting 50m to x

    t2 = 50/1.05
    t= 6.9 seconds

    For b.)

    Doing the same for finding the velocity of the truck above i would get

    3.4t = V (for the car)

    Also by computing

    [itex]\int[/itex]3.4t dt = [itex]\int[/itex]dx

    I would get

    1.7t2 = x

    Substituting 6.9 seconds would result to
    x = 80.94 m

    subtract this from 50m the

    80.94 - 50 = 30.94m

    Is this correct? Thanks in advanced :D
     
  2. jcsd
  3. Aug 12, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me!
     
  4. Aug 12, 2011 #3
    Whew, I was having doubts with my solution. Thanks!
     
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