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 Post subject: HELP!!! One more velocity problem!Posted: Sat, 12 Sep 2009 22:54:08 UTC
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Joined: Sat, 13 Sep 2008 00:23:07 UTC
Posts: 33
A ball is released from rest on an inclined segment of track (Track 1). The ball rolls down the incline onto a 4 m section of level track (Track 2) and the rolls back up Track 3, which is inclined at the same angle as Track 1. The ball finally rolls across Track 4, which is level.

I have a diagram showing that the ball starts at rest t = 0.0s and crosses Track 1 at t = 2 s. It starts Track 2 at t = 2s and finishes at t = 2.5s. It starts Track 3 at t = 2.5s and finishes at t = 3 s. Starts Track 4 at t = 3s and finishes at t = 4s.

A.) Find the uniform speed of the ball on Track.

This one I can do: 2m/2.5-2s = 2/0.5 = 4 m/s.

B.) Determine the length of Track 1.

I'm stuck here - I know it moves at a uniform speed of 4 m/s and clears Track 1 in 2 seconds. Do I do 4x2 = 8 m/s?

C.) Determine the acceleration, magnitude and direction, of the ball on Track 3.

Again, I'm stumped.

D.) Determine the length of Track 4.

Still stumped.

Help please?

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 Post subject: Posted: Mon, 14 Sep 2009 00:27:59 UTC
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Joined: Wed, 19 May 2004 22:26:01 UTC
Posts: 551
simplify and clarify the data

2 secs track 1

½ sec track 2 â€¦â€¦â€¦â€¦â€¦. 4m long

½ sec track 3

1 sec track 4

A)
Speed of ball on track 2 ........ distance/time = 4m/(1/2 sec) = 8m/s

B)
Speed at bottom of track 1 is 8m/s

Average speed = (0 + 8m/s)/2 = 4m/s

Length of track = ave. speed x time = 4m/s x 2sec = 8m

C)
the acceleration of the ball on track 3 is â€“ acceleration on track 1

acceleration of ball on track1 is (change of speed)/(time) = (8m/s)/2sec = 4m/s/s

acceleration of ball on track 3 is -4m/s/s

speed at end of track 3 = 8m/s + (-4m/s/s x ½ sec) = 6m/sec

Average speed = (8m/s + 6m/s)/2 = 7m/s

Length of track = ave. speed x time = 7m/s x 1/2sec = 3.5m

D)
Length of track 4 â€¦â€¦ distance = speed x time = 6m/sec x 1 sec = 6m

total length = 8m + 4m + 3.5m + 6m = 21.5m

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