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Homework Help: Question about accelerating movement

  1. Jun 29, 2008 #1
    A uniformly accelerating movement problem that we found in our book and we wanted to solve, but we are not sure if our solution is right:

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

    Two cars, (1) and (2), initially at rest, start to move from the same point, in the same road. Car (2), however, starts to move 3.0 seconds after car (1). The picture below (attached in .BMP format) represents, in Cartesian graph, how their scalar velocities vary in function of time during 18 seconds, counting from car (1).
    a) Calculate the scalar accelerations of cars (1) and (2) after their movements started.
    b) Verify if, until the instant t = 18 s, car (2) managed to reach car (1). Justify your answer.

    2. Relevant equations

    [tex]S = S_0 + v_0t + \frac{a}{2}t^2[/tex]
    [tex]a = \frac{\Delta v}{\Delta t}[/tex]

    3. The attempt at a solution

    The first question asks to find the accelerations.
    The accelerations, [tex]a_1[/tex] and [tex]a_2[/tex], are (calculating until instant t = 12, where both cars have velocity v = 18 m/s):
    Car (1): [tex]a_1 = \frac{\Delta v}{\Delta t} = \frac{18}{12} = \frac{3}{2} m/s^2[/tex]
    Car (2): [tex]a_2 = \frac{\Delta v}{\Delta t} = \frac{18}{12 - 3} = \frac{18}{9} = 2 m/s^2[/tex],
    noting that, since car (2) begins from [tex]t[/tex] = 3 seconds, it only moved during 12 - 3 = 9 seconds.
    We are sure this first part is right.
    Now, the second question is the one that generates a little doubt, since car (2) starts from the same point as car (1), but only 3 seconds after.
    Here is our attempt:
    First, we have to discover the space equation ([tex]S_1[/tex] and [tex]S_2[/tex]) for both cars.
    Car (1) is easy -> [tex]S_1 = \frac{a}{2}t^2[/tex], since it starts from rest and considering it starts from the position 0 meters. thus: [tex]S_1 = \frac{\frac{3}{2}}{2}t^2 = \frac{3}{4}t^2[/tex].
    Now, car (2), considering it started from time t = 0: [tex]S_2 = \frac{2}{2}t^2 = t^2[/tex]. Since it started moving from the same point as car (1), its initial position is also 0.
    To discover whether car (2) reached car (1) until t = 18 s, we have to know how much they moved until this instant. For car (1):
    [tex]S_1 = \frac{3}{4}t^2 = \frac{3}{4}18^2 = \frac{3}{4}324 = 243 m[/tex].
    For car (2), since it started at instant t = 3, we had to consider it only moved during 18 - 3 = 15 seconds. thus:
    [tex]S_2 = t^2 = 15^2 = 225 m[/tex].
    But the problem with the equation for car (2)'s movement is that it has to consider this car started its movement from time [tex]t[/tex] = 0.
    Therefore, the answer would be: no, car (2) didn't reach car (1) yet, because [tex]S_2[/tex] = 225 is smaller than [tex]S_1[/tex] = 243.
    Another conclusion that we drawed is that, if we wanted to discover the position where cars (1) and (2) meet, we should do [tex]S_1 = S_2[/tex]; but, since car (2) starts from a different time than car (1), we should do like this (in order to find time according to the referential presented in the graph, right?):
    [tex]\frac{3}{4}t^2 = (t - 3)^2[/tex].
    Does all this sound right?
    Thank you in advance.

    Attached Files:

    • pf.bmp
      File size:
      284.7 KB
  2. jcsd
  3. Jun 29, 2008 #2
    Hmm... your attachment is still pending approval. Could you meanwhile describe your graph? (as a piecewise function, etc.)
  4. Jun 29, 2008 #3
    We will try to describe:
    It is a velocity x time graph (with velocity as the y-axis and time as the x-axis).
    There are two functions described in the graph. They are straight lines, and encounter at a point, (t = 12; v = 18):
    Car (1) is a straight line from point (t = 0; v = 0) to point (t = 12; v = 18), and further prolonged until instant t = 18 (without specifying velocity at this instant).
    Car (2) is a straight line from point (t = 3; v = 0) to point (t = 12; v = 18), further prolonged until instant t = 18.
    Thus, the two straight lines start with 3 seconds of difference and cross at instant t = 12 (with corresponding velocity v = 18), and continue moving away from each other until instant t = 18.
    We are afraid that this description is confuse.
    Thank you in advance.
  5. Jun 29, 2008 #4

    Doc Al

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    Staff: Mentor

    Sounds good to me.
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