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Homework Help: A acceleration chase problem?

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data
    theres one thing i don't get about this problem

    "David is driving at a steady 30/s when he passes Linda, who is sitting in her car at rest. Linda then begins to accelerate at a steady 2m/s^2 at the instant David passes her"

    How far does tina drive before passing him?
    What is her speed as she passes him?

    2. Relevant equations
    Vf^2=vi^2 +2ad
    D= Vit +1/2at^2

    3. The attempt at a solution

    ok so i used a=v/t to solve for the amount of time it takes to catch up to david,i used 2=30/t where t would=15 seconds

    I then used that 15 seconds to calculate the distance Linda has travelled using the d=vit +1/2at^2 i got 225 from that

    and then i calculated her speed by using Vf^2=Vi^2 + 2ad which equaled 30m/s

    I'm just not sure about that first step when i found the time, is that part right? i'm pretty sure my last 2 steps were fine, but is that 2m/s^2=30/t right?, any help would be appreciated
  2. jcsd
  3. Sep 7, 2011 #2


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    No! Since the time David passed Linda, David has been traveling at 30m/s. Linda has always been traveling slower than David, until Linda hits 30m/s. How can that be the moment she passes him? She must be behind him at that point. Compare the distance traveled at a steady 30m/s with the distance traveled starting from rest at an acceleration of 2m/s^2 when they are equal is when they pass. And who is Tina?
  4. Sep 7, 2011 #3
    i see your point, but how are you supposed to get the distance if we only have 1 variable for david? do i have to calculate it in conjunction with Linda's acceleration? but i'm just confused on which equation you would use for that.
  5. Sep 7, 2011 #4


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    When Linda passes David, their positions are the same. So, write an expression for the position vs. time of David, write another expression for the position vs. time of Linda, and equate them. Solve for t to get the time at which Linda's position is equal to that of David's.

    It's probably easiest if you take t = 0 to be the moment at which David passes Linda and she begins accelerating from rest. It's probably also easiest to call that position 0 and measure all distances from it.
  6. Sep 7, 2011 #5


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    Try your third equation. Linda starts from rest and accelerates at 2m/s^2. How far does she go in time t? David travels at a constant velocity of 30m/s. How far does he go in time t? For what value of t are the two distances equal? That's when they pass each other, right?
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