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Homework Help: Linear Motion (I'm sure it is easy for all you majors out there)

  1. Nov 21, 2005 #1
    Linear Motion (I'm sure it is easy for all you out there)

    Okay, so I'm not really getting this problem:
    "A bowling ball traveling at constant speed hits the pins at the end of the lane, 20m away. If the bowler hears the "crack" of the ball hitting the pins 2.5 s after releasing the ball. If the speed of sound is 340m/s, what is the speed of the ball?"
    My first thought is to use distance over time equals velocity (d/t=v)
    So I thought that I might try 20m/2.5s, but that only tells me the velocity of a ball going down the lane... I'm not really sure how to incorporate the speed of sound within the 2.5 secs...
    My next "problem" is: "A speeding motorist passes a stationary police officer at 120 km/h. If the officer immediately accelerates at 5 km/h/sec, how long before the officer catches the speeding motorist?"
    For some reason my instinct in this situation is to go 120x = .5(5km\h\sec)(x)^2, but it just doesn't work...
    So I thought that perhaps I could try converting the 120km/h to km/h/s: 120km/h/3600sec = .033km/h/s and I wanted to use x again, but I'm just at a loss of words...
    Maybe I am biting off more than I can chew, but I am trying.
    Thanks in advance for any help.
    Last edited: Nov 21, 2005
  2. jcsd
  3. Nov 21, 2005 #2
    I think your first question is hinting that the time it takes the CRACK to reach your ears is after the collision because sound travels at (believe it or not) the speed of sound.

    So think about:
    How long does it take sound to travel 20m?
    Then you can work out how long it took the bowling ball to travel 20m.
    Then you can work out the speed of the ball.

    Doesn't it? I think it will, but remember (with these units), you will get an answer is seconds.

    Does this help?

  4. Nov 21, 2005 #3
    Okay, so I was on the right-track on the second-one!!
    But... The first one still has me :yuck:
    In this problem I know it takes it d = at, or 20m = 340m\s * t
    That would make .0588 = t, so it takes .0588 seconds for the speed to travel... 2.5s - .0588 = 2.441seconds left
    .. Okay, so the ball went 20meters in 2.44seconds... v = d/t, 20m/2.44s = 8.196m/s, the the answer is 8.19m/s!! :surprised: Sweet. Maybe I'm not hopless afterall...
    There is one more problem I could use help with on this linear motion stuff :bugeye:
    "You have created a catapult to launch your friend onto the school roof. At what vertical initial velocity must they be going in order to make it up to the 5 meter high roof?"
    | 5 m
    ----- is how I picture it...
    So I want to say the equation I am solving is: d = vt;
    So I want to figure out the time since I know the distance: 5m = .5(9.8m\s)t^2, atleast I think the deaccleration would be 9.8m\s since the ball is going up and is loosing inertia due to g, right? Or at a rate of g?
    Anyway, 9.8*.5 = 4.9m\s(t^2) = 5 meters
    5/4.9 = 1.02 = t^2
    Sqroot of both sides...
    1.01 = t
    According to this t is 1.01 seconds, so if I plug that in :
    5 meters = v(1.01)
    Makes v = 4.95m\s, where I know the answer isn't that... Any help in where I am fowling up?
  5. Nov 23, 2005 #4
    Actually, it looks like what I am trying to solve is d = vi(t) + .5(a)t^2...

    Accleration would seem to me to be 9.8m\s...

    5m = vi(t) + .5(9.8)t^2

    Grr, but how to solve for t?

    d/t=v ?
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