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Homework Help: Centripetal and Tangential Acceleration (Change in Speed over time)

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data
    A car is rounding a curve on the interstate, slowing from 30 m/s to 22 m/s in 7.0 seconds. The radius of the curve is 30 meters. What is the acceleration of the car?
    vi=30 m/s
    vf=22 m/s
    t=7 s
    r=30 meters.

    2. Relevant equations
    change in velocity/change in time= tangential acceleration ?
    centripetal acceleration = v^2/ r = -radial acceleration

    a= a_radial + a_tangential (acceleration in vector notation)

    3. The attempt at a solution

    change in v/ change in time= (22m/s-30 m/s) / (7s)= tangential acceleration= -1.14 m/s^2

    I'm not sure how to find the centripetal acceleration? Does the problem mean find the total acceleration at each point? Meaning using each v to find centripetal acceleration?

    (30)^2/ 30=
    (22)^2/ 30=

    Maybe the difference between the two?
    There's no way to check my answer.
  2. jcsd
  3. Sep 5, 2011 #2


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    welcome to pf!

    hi kgarcia! welcome to pf! :wink:
    it's a poorly worded question :frown:

    yes, it must mean the total acceleration at each point, since the https://www.physicsforums.com/library.php?do=view_item&itemid=27" will be different at different points, depending on v (though the linear acceleration is presumably constant) :smile:
    Last edited by a moderator: Apr 26, 2017
  4. Sep 5, 2011 #3
    if you took calculus, you'd know that if you drew a line through the 22m/s and 30m/s a secant line would form. the question you are faced with would then be how to find the tangent of this. it would be the mean value theorem I think. so mean value theorem is like s(b)-s(a)/b-a and in this case is V(7)-V(0)/7seconds which tells you acceleration of time. that's one way of looking at it.

    hm....this is a weird problem.

    so if a(t)=(vt)^2/r then (vt)^2/30=a(t) and the only problem is vt. I'm sure the acceleration is constant, but for some reason when you plug in for vt, the centripetal acceleration is different at each tengential velocity. but if they ask for the acceleration of the car, does that necessarily mean centripetal acceleration? Well They ask what "IS" the acceleration of the car, which may refer to the 30m/s by itself. But Im almost sure that you were right the whole time.
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