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Is this correct without a direction in metres?

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data

    A rocket-powered hockey puck moves on a horizontal frictionless table. Figure EX4.8 at the top of the next column shows graphs of vx and vy, the x- and y-components of the puck's velocity. The puck starts at the origin.

    a. In which direction is the puck moving at t = 2s? Give your answer as an angle from the x-axis.
    b. How far from the origin is the puck at t = 5s?

    The graph of the X-velocity is v=8t cm/s and the Y-velocity is V=30cm/s

    2. Relevant equations

    c^2 = a^2 + b^2
    d = vt
    m= dy/dx

    3. The attempt at a solution

    a. vx = 16 cm/s. vy = 30cm/s. θ = tan (30/16)^-1 = 62^o. Therefore the puck is traveling in the direction of 62^0 from the x-axis.

    b. vx = 8* 5 = 40cm/s
    dx = 40*5 = 200cm

    dy= 30*5 = 150cm

    c = (4000+2250)^0.5 = 250cm

    Therefore, the puck is 250 cm from the origin.

    My question are these:

    For a I do not need to add the displacement of it, correct? Just the angle is good enough right? In high school there was never a question that did not involve the magnitude of the displacement.

    Secondly. Is the position traveled from the origin correct as well? I am asking this because I am a bit confused because the position function for x-direction is 4x^2. It shouldn't matter if it is linear or not correct. I can still do it the way I solved it correct? As well for b, I do not need to include the angle because it is asking for the magnitude of the displacement, correct?
     
  2. jcsd
  3. Sep 30, 2011 #2

    NascentOxygen

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    Staff: Mentor

    Your (a) looks correct.

    200 cm is the distance it would have covered were it travelling at a constant speed of 40cm/s since the start. It has been travelling at less than this all the while, so your answer cannot be correct.

    You need to find the displacement (from the origin) at time t=5 secs. The magnitude of this gives you the distance from the origin at t=5. Displacement is found as the area under the v-t graph. You have the graph, so you can work out the area under it.

    Area has units. What units will the area under your v-t graph have?
     
  4. Sep 30, 2011 #3
    Thanks, I forgot to half the area found because it was a triangle. To prove me correct I took the integral. I got 100 cm from x. So the final answer is 180.27cm from the origin, correct?
     
  5. Sep 30, 2011 #4

    NascentOxygen

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    Staff: Mentor

    My answer is 180.27cm also.
     
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