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Homework Help: Constant velocity and acceleration problem

  1. Oct 18, 2006 #1
    having trouble setting these style of problems up. please take a look.

    problem: A student it running at a constant velocity of 6.00m/s to catch the campus shuttle bus, which is stopped at the bus stop. When the student is 80 m from the bus, it starts to pull away with a constant acceleration of .200 m/s^2.

    (a) How far does the bus travel in the first 10.0s?
    (b) How far has the bus traveled when its speed is 1.75 m/s?
    (c) From the instant when the bus starts accelerating, how long will the student have to run before catching the bus?
    (d) How fast will the bus be traveling when the student catches the bus?
    (e) The equation you solved in (c) has a second solution. What is the significance of that solution?
    (f) what is the slowest speed that the student can have and still catch the bus?


    a. == 10.0 m
    b == 7.66m
    c == 20.0 s or 40.0 s
    d == 4.00 m/s
    e == The student first catches up with the bus at t=20.0s. If she keeps running, she then passes the bus. The bus continues to accelerate and passes the student when t = 40.0s
    f == 5.66 m/s


    what i've done.

    i've used formulas to try to solve these:
    x = x0 +v0t + 1/2at^2
    and ... v = v0 +at
    and ... v^2 = v0 + 2a*(chng Of X)

    however, i don't get the right answers, i know to get the right answers i can't mix match the information in the formula.

    okay for a. and b. i get the correct answers can you please verify my work is correct. and i need help setting up the rest?

    for a, i first found velocity: v = 0 + (.2)(10s) = 2 m/s.
    then used a different formula to find the distance for questions a.

    so x = 0 + 0(10) + 1/2(.2)(10^2) = 10 m.

    alright for b. i just used the third formula: (1.75)^2 = 0 + 2(.2)(chgofx) === 7.656

    for c,d,e,g and f i get confused. please help!
  2. jcsd
  3. Oct 18, 2006 #2


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    For c you need to write the equation of motion for the student. It has exactly the same form as the equation you used for the bus. There are a couple of ways you could do this, but a good way is to say that x = 0 for the student at the point where the bus starts. What does that make x0 for the student? What is his v0? What is his a? If you do it this way, the x for the student must equal the x for the bus when he catches up, and the times must also be the same for both.

    Once you have the time from c you can do d

    For f, the student has to catch the bus (same x values) before the bus velocity is the same as the student velocity. You can figure out how much time that will take and work from there.

    Note: You have an error in your third formula. The v0 should be squared.
  4. Oct 19, 2006 #3
    ya formula was wrong there, thanks.

    okay i'm still confused how you can get all the information..

    if x = 0, then what would x0 =?

    so when he's 80 meters from the bus, x = 0?
    btw, doesn't x stand for his final position so how could it equal 0?

    also, since hes already running, is constant velocity the v0? his beginning speed? or do we always assume something is at rest at v0?

    i think i can do this: 6.0m/s = 0 +a(10) == .6 m/s^2
    but why is t = 10?
    we just assume that in a 10 second measure of time, this is whats happening? and how can i know t, if thats the unknown were trying to find?

  5. Oct 19, 2006 #4


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    I think you are not seeing the situation clearly.

    You have two things moving in the same direction, starting in different places, with different speeds, and different accelerations. If you choose the zero of time to be when the bus starts accelerating, then the position and velocity of the bus are given by of the bus is given by

    xb = xb0 + vb0t +(1/2)abt^2
    vb = vb0 + abt

    Using the given information, you can set xb0 = 0 and vb0 = 0 to reduce this to

    xb = (1/2)abt^2
    vb = abt

    t = 0 has to be when the acceleration starts for the equation to be valid.

    The student has similar equations

    xs = xs0 + vs0t +(1/2)ast^2
    vs = vs0 + ast

    but the student starts in a different place (80m from the bus) with a different initial velocity (6m/s) and no acceleration. His equations reduce to

    xs = xs0 + vs0t
    vs = vs0

    At t = 0 the student is 80m behind the bus, so xs0 = ???
    This answer depends on whether you want to use the same coordinate system for the student as for the bus, or two different coordinate systems. If you use the same one, then you know that when the student catches the bus they have the same x value at the same time. You could choose a different coordinate system for the student; the important thing is that the student has to go 80m more than the bus to catch up. I suggest you use the same coordinte system, which means xs = 0 when the student gets to the point where the bus started. So what must xs0 be?

    You have all the other values to put into the equations. The equations can be solved for the position and time when xb = xs, which is where and when the student catches the bus. Here is a little diagram of the situation showing the positions of the bus and the student at t = 0 and when the student catches the bus, where the choice has been made to call the initial position of the bus x = 0. The bus and the student are moving to the right.

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