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Acceleration on a curve [kinematics]

  1. May 21, 2009 #1
    1. The problem statement, all variables and given/known data

    a car is moving forward at 60 km/h [E 45° N], executes a turn and is now moving at 35 km/h [N 80° W]. If the turn takes 8 seconds to complete the turn what is the average acceleration of the car in m/s?

    2. Relevant equations

    v2 = v1 + a(t)

    3. The attempt at a solution

    v2 = 35
    v1 = 60
    t = 8
    a = ?

    v2 = v1+ a(t)

    35km/h = 9.72 m/s
    60km/h = 16.6 m/s

    35 = 60 + a(8)
    35-60 = 8a
    a= -3.125m/s

    i'm not sure if its right, because don't you accelerate positively when you change direction
    Last edited: May 21, 2009
  2. jcsd
  3. May 21, 2009 #2


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    Homework Helper

    Welcome to PF.

    Remember velocity is a vector. As is acceleration. So ... that means that you might do better to resolve the velocities into their components and then figure the change in velocity by x,y to determine the direction and angle of the average change over the 8 seconds.
  4. May 21, 2009 #3
    ok well based on what you said: split into x and y and find average i came up with this diagram:

    http://img132.imageshack.us/img132/3949/diagram.jpg [Broken]

    I used the previous values (60), (35) to create the x and y for the purple triangle and finally finding the average speed represented by the "?", i then divided that value by the time (8 seconds) and got my answer. Was my method correct?
    Last edited by a moderator: May 4, 2017
  5. May 21, 2009 #4


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    Homework Helper

    I prefer to do it algebraically.

    Vi = Vix + Viy

    Vi = |Vi|*Cosθ x + |Vi|*Sinθ y
    Last edited by a moderator: May 4, 2017
  6. May 21, 2009 #5


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    Homework Helper

    As to your drawing ...

    You want to keep in mind that

    a = ΔV/Δt


    ΔV = Vf - Vi
  7. May 21, 2009 #6
    alright thank you, solved :D
  8. May 21, 2009 #7
    Hi guys,

    I have similar problem with boogaaaa.

    As we know, if a car accelerate on curve at certain speed, it will tilt outside the circle track. So i sketch this 2 picture to show how it works.

    http://img193.imageshack.us/img193/2323/66057246.jpg [Broken]
    http://img194.imageshack.us/img194/9272/96495741.jpg [Broken]

    I just wondering, how can the car tilt or flip outside the circle track if the side force (centripetal force) is pointing inside the circle?

    My suggestion:
    Total Moment about tyre (2); ∑M2=0

    So, Fs*h + G*b/2 - FN1*b = 0

    I assume that FN1=0 since the car is about to tilt. So here come the problem.

    Fs = - G*b / 2h.

    The minus sign shows that the side force, Fs, should point outside the circle.

    can anyone help me?
    Last edited by a moderator: May 4, 2017
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