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Homework Help: Seemingly easy mechanics problem (constant acceleration)

  1. Aug 31, 2013 #1
    1. The problem statement, all variables and given/known data
    A jet plane comes in for a landing with a speed of
    100 m/s, and its acceleration can have a maximum magnitude
    of 5.00 m/s^2 as it comes to rest. (a) From the
    instant the plane touches the runway, what is the minimum
    time interval needed before it can come to rest?

    2. Relevant equations
    Kinematics I suppose, but I'm trying to solve it with just calculus.

    3. The attempt at a solution

    So my idea was that if the acceleration is constant at -5m/s^2 then I should be able to take the integral of this and get the velocity function. So what I end up with is

    v(t) = - 5/2 (t^2) + 100

    Solve for v(t) = 0 etc.

    But the answer is wrong. It's supposed to be 20s.

    Where is my logic flawed?
  2. jcsd
  3. Aug 31, 2013 #2


    User Avatar

    Staff: Mentor

    Check where that power of 2 came from.

    You should know a family of constant acceleration formulae, including
    v = u + at
  4. Aug 31, 2013 #3
    I made it up as I took the integral. I'm not sure what you're getting at here. What do you mean?
  5. Aug 31, 2013 #4
    There are a few equations in mechanics that are linear and deal with constant acceleration.

    V = u + at

    is one of them

    U is initial velocity
    V is final velocity
    a is acceleration
    t is time

    If you rearrange the formula, you can work out the time taken.
  6. Aug 31, 2013 #5
    With just calculus you just got wrong the formula of v. You have a= -2.5 . When you integrate that you have??? not t^2 obviously...
  7. Aug 31, 2013 #6
    That's the thing, I wasn't using the kinematic equations. It seems to me that I should be able to solve this using just calculus. My reasoning was that you can solve "gravity problems" that way. If you integrate -9.8t you get the velocity function for dropping something, -4.9t^2, correct?
    Then why doesn't this work?

    I integrated a= -5t not -2.5t^2.
  8. Aug 31, 2013 #7
    Yes when you integrate a = dv/dt = -5 !!! you get v = -5t + v0 (=100) not v= -(5/2)t +100. ;)
  9. Aug 31, 2013 #8

    Doc Al

    User Avatar

    Staff: Mentor

    The acceleration is -5, not -5t. It's constant.
  10. Aug 31, 2013 #9
    Ohhh my gosh! Of course! Thank you so much Doc Al, you saved my day. That makes complete sense.

    I lift my hat for you Sir!
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