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Regarding Forces & Motion

  1. Oct 29, 2007 #1
    Regarding Forces & Motion (edited)

    Hi, thank you for your time to read this thread.
    I'll begin with the question, thanks again!

    Q: At the instant a traffic light turns green, a car starts from the rest and accelerates
    uniformly at a rate of 4.0 m/s[tex]^{2}[/tex] East. At the same instant, a truck traveling with a
    constant velocity of 90.0 km/h East overtakes and passes the car.

    a) How far beyond the starting point is the car after 10.0 s?
    b) How far beyond the starting point is the truck after 10.0 s?
    c) The car passes the truck at a distance of 312.5m beyond the starting point.
    How fast is the car travelling at this instant?
    d) How long does the car take to pass the truck?
    e) Both drivers suddenly see a barrier 100.0m away and hit their brakes at exactly
    same time. Assuming that both vehicles decelerate uniformly and they take 3.0 s
    to stop, will the stop in time?

    Here are my answers :)
    A: (I've omit vector signs, not sure how to input the symbol)
    a) a = 4.0 m/s[tex]^{2}[/tex], [tex]\Delta[/tex]t = 10.0 s
    [tex]\Delta[/tex]d = (v)(t) + 1/2(a)(t^2)
    [tex]\Delta[/tex]d = (0)(10) + 1/2(4)(100)
    [tex]\Delta[/tex]d = 200m

    b) v = 90.0 km/h (convert to m/s which equals to 25m/s), [tex]\Delta[/tex]t = 10.0 s
    [tex]\Delta[/tex]d = (v)(t)
    [tex]\Delta[/tex]d = (25)(10.0)
    [tex]\Delta[/tex]d = 250m

    c) [tex]\Delta[/tex]d = 312.5m, a = 4.0 m/s[tex]^{2}[/tex]
    vii[tex]^{2}[/tex] = vi[tex]^{2}[/tex] + 2a[tex]\Delta[/tex]d
    vii = (0) + 2(4.0m/s[tex]^{2}[/tex])(312.5m/s)
    vii = 50 m/s

    d) a = 4.0 m/s[tex]^{2}[/tex], vii = 50 m/s
    [tex]\Delta[/tex]t = (vii - vi) / a
    [tex]\Delta[/tex]t = (50m/s - 0) / 4.0 = 12.5 seconds

    e) Here, what I thought of doing was, find both car and truck's acceleration
    then find their distance within 3 seconds then check whether the answer is
    higher than 100m or not.

    a = [tex]\Delta[/tex]v / t
    a = 0 m/s - 50 m/s / 3.0s
    a = -16.7 m/s[tex]^{2}[/tex]

    [tex]\Delta[/tex]d = (vii)[tex]\Delta[/tex]t - 1/2a[tex]\Delta[/tex]t[tex]^{2}[/tex]
    [tex]\Delta[/tex]d = 75.15m

    Same calculation
    [tex]\Delta[/tex]d = 37.35m

    // So I've said that those two vehicles will stop on time because the calculated
    distance is less than 100m.


    Thank you for reading this.. I really appreciate it.. It's very long..
    I needed some advice especially on e), because that question really got me thinking.
    I honestly don't have confidence that my answers are right :( If I'm wrong
    please correct me XD

    Thanks again!
    Last edited: Oct 29, 2007
  2. jcsd
  3. Oct 29, 2007 #2


    User Avatar
    Gold Member

    a) It may be using complex math, but otherwise no.
    how about using the formula [tex] s=v_i t+\frac{at^2}{2}[/tex]

    b)You know it's velocity, which is constant, so how would you calculate distance over a certain time interval?

    Remember to use SI units.
  4. Oct 29, 2007 #3
    So the V in that equation represents the initial, which is 0 right? t would be the given unit
    in the question and same for b)

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