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Need help with 2 probelms - 1 simple one

  1. Aug 31, 2006 #1
    need help with 2 probelms - 1 "simple" one

    1) a 50.0 g superball traveling at 25.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. a high-speed camera records this event. if the ball is in contact with the wall for 3.50 ms, what is the magnitude of the average acceleration of the ball during this time interval? (note: 1ms = 10^-3s)

    heres how I tried to do this problem. first, I converted 3.50 ms to seconds which is .0035 seconds. Im assuming this will be my change in time.

    now, avg acceleration = change in velocity / change in time

    so heres how i tried to get change in velocity. since the final velocity of the ball right before it hit the wall is -25 m/s (Vf1) [negative because its going towards the wall] , that would then become the Initial velocity when it bounces off the wall(Vi2).....and so Vf2 is +22.0 m/s. [positive since its going away from the wall]. is my logic here right?

    avg accel = +22 - (-25)/.0035 = 13428.57m/s^2!!!!!! does that not seem right at all!!!

    2) a 745i BMW car can brake to a stop in a distance of 121 ft. from a speed of 60 mi/h. to brake to a stop from a speed of 80 mi/h requires a stopping distance of 211 ft. what is the average braking acceleration for a) 60mi/h to rest b) 80 mi/h to rest c) 80 mi/h to 60 mi/h.

    when they say average braking acceleration.....do they mean "average acceleration" like the problem above? I figure its something more complicated than that, can someone point me in the right direction for this?
     
  2. jcsd
  3. Aug 31, 2006 #2

    Andrew Mason

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    You want to find the average acceleration. a = dv/dt. The force is ma= mdv/dt. This is just finding the change in momentum, which is equal to the Force x time. [itex]F = dp/dt = m\Delta v/\Delta t[/itex].

    They mean average rate of change of speed: [itex]a = \Delta v/\Delta t[/itex]
    Use [itex]d = v_{avg}t = \frac{1}{2}v_0t[/itex] to determine the time it takes to stop and then find a.

    AM
     
  4. Aug 31, 2006 #3
    It's right, but you have to remember that that acceleration is happening for only 3.50ms, or 0.0035 ms!! So even though the acceleration is very high it only actually acts on the ball extremely momentarily. (Also make sure to mention that you chose the way the ball bounced as positive [and hence the original direction negative])
    --
    Edit: You can use derivatives etc. but you really don't have to, I did physics a year early and hadn't learnt derivatives and I went though fine (assuming that outside Australia the standard is similar).
    --
    Breaking acceleration is just a way of saying deceleration, or retardation (or acceleration in the opposite direction of movement). I think they're trying to test your ability to interperate questions (or confuse you). You're going to have to do conversions of units here, as well as use u v and x (distance) to find t.
     
    Last edited: Aug 31, 2006
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