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Combining two equations

  1. Oct 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Bob is running at a constant speed of 5.00 m/s when Alice runs past him in the same direction. Precisely 1.00 s after Alice passes him, Bob starts to accelerate at a constant rate of 2.00 m/s2. Bob catches up with Alice after accelerating for 6.00 s. Alice kept the same constant speed the entire time. What was her speed?

    2. Relevant equations
    vf=vi + at

    3. The attempt at a solution
    Since Alice is running at a constant velocity the whole time, her a=0 m/s^2. That means her initial velocity is equal to her final velocity.

    vf = vi

    Now for Bob:

    vf = vi + at
    vf = (5m/s) + (2 m/s^2)(6s-1s)
    vf= 15 m/s

    So Alice's velocity must have been 15 m/s. I know I calculated Bob's velocity. This is not the correct answer. What am I doing wrong?
     
  2. jcsd
  3. Oct 23, 2015 #2

    SteamKing

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    It's not enough that Bob matches Sally's speed in order to catch up. When Bob catches up with Alice, their locations must coincide. What equations must you use in order to determine where Bob or Sally is at any given time?
     
  4. Oct 23, 2015 #3
    x = x0 +v0 + (1/2)at^2.

    So if both of their locations must equal to each other, then:

    Alice:
    x = x0 + v0t + (1/2)(0)t^2
    x = x0 + v0t

    Bob:
    x = x0 +v0t + (1/2)at^2

    If we combine these two together:

    v0t = v0t + (1/2)at^2
    v0(5s) = (5 m/s)(5s) + (1/2)(2m/s^2)(5s)^2
    v0(5s) = 25m + 25m
    v0(5s) = 50m

    Alice's velocity = 10 m/s

    Is this right?
     
  5. Oct 23, 2015 #4

    SteamKing

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    Does your equation for the distance Bob ran after Alice match the description from the original problem statement? To wit:

    From the Problem Statement:
    1. Bob is running at 5 m/s, constant velocity.
    2. 1 second after Alice passes Bob, Bob accelerates at 2 m/s2.
    3. After accelerating for 6 seconds, Bob catches Alice.

    From Your Equation:
    1. Bob is running at 5 m/s, constant velocity.
    2. Bob accelerates at 2 m/s2.
    3. After accelerating for 5 seconds, Bob catches Alice.

    I think there are some discrepancies between the two descriptions here.
     
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