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Homework Help: Acceleration in a joy ride

  1. Sep 18, 2010 #1
    i got this question in my first assignment in PHYS 109. i am sure the question is easy, but for some reason i can't get it around my head.
    Thanks for the help !!

    1. The problem statement, all variables and given/known data

    Code (Text):
    A physics 109 student is taking a car for a joy ride on a long, empty and perfectly
    straight Saskatchewan highway. At the beginning of the joy ride the student has the
    car accelerate from rest to some top speed with a (constant) acceleration of 4:50 m=s2
    and then maintains that top speed to the end of the ride. The total time taken from
    the start of the joy ride to its end is 1:36 minutes and the total distance covered is
    3:84 km.
    (a) What was the time elapsed during the acceleration phase of the car's motion?
    (b) What distance did the car cover during the acceleration phase of its motion?
    (c) What was the top speed reached by the car?

    2. Relevant equations

    V = Vi +at
    Xf=Xi +Vi(t) + (1/2)(a)(t^2)

    3. The attempt at a solution

    t=81.6 s
    Xf =3.84 * 10^3 m

    a) Vavg. = X/t = 3480/81.6 = 47.1 m/s

    Vavg= (Vf + Vi)/2
    => Vf= 2(Vavg - Vi)
    ==>Vf= 2(47.1 -0)= 94.1 m/s

    Vf=at +Vi
    =>t=(Vf - Vi)/a = 20.9 s

    Xf=Xi +Vi(t) + (1/2)(a)(t^2)

    Xf= 0+0+(0.5)(4.50)(20.9^2) = 982.8m

    C)Vf = 94.1 m/s

    i dont belive in my answer because :
    1) The top speed is 339 KM/h....yes it is possible, but he must be one rich Phys student we are talking about here

    2) I got the Vf in the very bigening, yet he asks me about it at the end ? then i must be doing somethiong wrong

    Thank you guys, anhy help is really appreciated
  2. jcsd
  3. Sep 18, 2010 #2
    Hi mspike

    You can't use x/t to find the speed because there is acceleration.

    Now consider two cases:
    1. when the car is accelerating
    2. when the car moves with constant speed after accelerating

    t1+t2 = 81.6 s
    s1+s2 = 3.84 x 103m

    Find s1 (distance traveled when accelerating) in term of t1, then find also s2 (distance traveled with constant speed) in term of t1.
    Use s1+s2 = 3.84 x 103m, then you'll get quadratic equation in term of t1. By solving that, you'll get two values for t1; reject the impossible one
  4. Sep 18, 2010 #3
    First of all thank you for your replay.

    i understood what you mean, but i dont know how to actually do it (i have been tryiong since you replaied)

    i dont know how to "Find s1 (distance traveled when accelerating) in term of t1, then find also s2 (distance traveled with constant speed) in term of t1. "

    the closed thing i could get to is.

    T1 + T2 = 81.6
    T1= 81.6 - T2

    S1f = S1i + V1i (T1) +(o.5)(a)(T1)^2

    S1f = 0 + 0 + (2.25)(81.6 - T2)^2
    S1f = (2.25) ( 6658.56 - 163.2 T2 + T2^2)

    You see where am going ? but, then the final answers are all impossible.

    if you could give me 1 or 2 more hints.

  5. Sep 18, 2010 #4
    Actually you expressed s1 in term of t2 (see your last equation :smile:)

    It also works. Now state s2 in term of t2. What is the formula to find s2?
  6. Sep 18, 2010 #5
    I would say the same equation as before

    S2f = S2i + V2i (T2) +(1/2)(a)(T2)^2

    S2f = S2i + V2i (T2) + 0

    V2i = V1f

    Dead end .

    Is there a better formula to use ?
  7. Sep 18, 2010 #6
    You are on the right track. Note that S2i = 0. Can you find V1f in term of t2?
  8. Sep 18, 2010 #7
    V1f = V2i

    V1f = aT1 +V1i

    V1f = 4.50 ( 81.6 - T2) +0
    V1f = 367.2 - 4.50 T2

    Since V1f = V2i

    Therefor, S2f = S2i + (367.2 - 4.50 T2) (T2)

    S2f = 0 + 367.2 T2 - 4.50 T2 ^2 *Shouldn't S2i be equal to S1f ? but am takeing your word for it*

    S1 + S2 = 3840

    (T22 -367.2 T2 + 14981.85 + 367.2 T2 - 4.50T22 = 3480

    -4.50T22 + 0 T2 + 14981.85-3840 = 0

    T2= 50.6 s

  9. Sep 18, 2010 #8
    After finishing the calculation i think there is still something wrong.

    If T2 = 50,6s then T1 = 31s

    Than to Calculate S1f we will say

    S1f = S1i + V1i + (1/2)(a)(T1^2)

    S1f = 2162.25 m (which makes no sense.. but still lets carry on)

    to Find Top speed (V1f)

    V1f2 = V1i2 + 2 (Delta x)a

    Which is equal to 139.5 m/s these guys must have one heck of a car ! :D
  10. Sep 19, 2010 #9
    No, S2i does not equal to S1f. We have broken the problem into two cases. So, the initial distance for each case is zero.

    Check the s1 again :smile:
  11. Sep 21, 2010 #10
    Thank you so much for your help Songoku !!
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