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Kinematics - calculating acceleration

  1. Mar 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Jack is driving with a pail of water along a straight pathway at a steady 25 m/s when he passes Jill who is parked in her minivan waiting for him. When Jack is beside Jill, she begins accelerating at the rate of 4.0 x 10-3 m/s2 in the same direction that Jack is driving. How long does it take Jill to catch up to Jack?

    2. Relevant equations



    3. The attempt at a solution

    I'm superbly and completely stumped by this, I really have no idea how to even begin solving this. If anyone could help that would be most appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 13, 2012 #2
    If Jack is driving at a steady 25 m/s, then kinematics tell us that:

    x(t)=25t

    Now, Jill, initially at rest, starts accelerating at the rate of 4.0E-3 m/s2, which results in:

    x(t)=(1/2)(4.0E-3)t2

    We are admitting that they both start their movement at the origin of (Oxy). You can now solve it for the same position (same x).
     
  4. Mar 13, 2012 #3
    so 25 = 2x10^-3 (t)^2
    25= SQRT0.002
    25= 0.0447
    t= 25/0.0447
    t= 559s
    or t = 9.3 minutes
     
  5. Mar 13, 2012 #4
    25t, not 25. You must solve a quadratic equation.
     
  6. Mar 13, 2012 #5
    you lost me sorry :'(
     
  7. Mar 13, 2012 #6
    These are the two equations. You forgot take the t term in the first equation into account.
     
  8. Mar 13, 2012 #7
    so how about...
    25 t^2 = .002 m/s^2 t^2 divide both sides by t^2
    25 = .002 m/s^2 divide both by 0.002^2
    so 25/ 0.002
    so if we cancel out the m/s we are left with a sec so an answer of time of 12500 seconds or 3.47 hours.
     
  9. Mar 13, 2012 #8
    Nice work, that is correct.
     
  10. Mar 14, 2012 #9
    It's partiallly wrong, Jack's speed is constant (thus no acceleration!).

    25t=(1/2)(4.0E-3)t2 → (2.0E-3)t2-25t=0

    You can either use the quadratic formula or cancel the product like so:

    t=0 [itex]\vee[/itex] t=25/(2.0E-3)=12500s

    You got the solution right but you used a wrong method as 25≠0.002.
     
  11. Mar 14, 2012 #10
    Wow whoops, totally didn't see that xD
     
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