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What is the speed of the ball 10.0 sec later?

  • Thread starter Peterson
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  • #1
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INTRODUCTION:
This is a problem from my Introduction to Physical Science class using "Conceptual Physics" 10th Ed.by Paul G. Hewitt

EXACT PROBLEM:
"A ball rolls down an inclined plane with a uniform acceleration of 1.00 m/sec^2. At a certain instant, its speed is 0.5 m/sec^2."

PROBLEMS FACED:
a) What is the speed of the ball 10.0 sec later?
b) How far has the ball traveled in that 10.0 sec time interval?
c) How far did the ball travel during the eighth second after it was released?

MY THOUGHTS:
I don't know what to do here really. I've had phyisics before, but that was for years ago (so much for saving that easy intro science class for senior year:rolleyes:). I am pretty much at a loss here. Any help for a lazy, "I just wanna graduate" senior would be of great help.
 
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Answers and Replies

  • #2
hage567
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Start by looking up the basic kinematic equations of linear motion in your book. Write out the things you know and the things you need to find. Find the equation that fits them together. Drawing a picture and labelling the points with the things you know can help.
 
  • #3
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Start by looking up the basic kinematic equations of linear motion in your book. Write out the things you know and the things you need to find. Find the equation that fits them together. Drawing a picture and labeling the points with the things you know can help.
That doesn't help me at all. I take it I'm looking for something dealing with the equations for speed, average speed, acceleration. I honestly don't know.
 
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  • #6
hage567
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Does your book not talk about them? I'm not trying to be difficult, but I find it hard to believe there aren't any examples of how to use these equations in your textbook.

To give you a hint for a):

A ball is travelling down an incline, with an acceleration (a) of 1 m/s^2. At some time, it has a velocity ( call it Vo) of 0.5 m/s. At t = 10s after it had the velocity of 0.5 m/s, you want to know what the new velocity (V) is. So you know Vo, a, t, and want to find V. Which one of those equations listed in the formulary do you think you could use?
 
  • #7
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ok, you answered my biggest problem. 0.5 is the starting point.
 
  • #8
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OK, so here's my crack at part a:

v = Vo + at
v = .5 + (1.0)(10)
v = .5 + 10
v = 10.5 m/s

part b:

d = vt
d = (10.5)(10)
d = 105 m

part c:

v = Vo + at
v = .5 + (1.0)(8)
v = .5 + 8
v = 8.5 m/s

d = vt
d = (8.5)(8)
d = 68 m

now was I correct in having to redo the velocity for part c?
 
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  • #9
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Since the ball is rolling wouldn't we have to use kinematic equations for uniformly accelerated rotational motion rather than linear motion (since the ball isn't sliding down the plane). So in terms of x, v, and a it's just a small gripe but use theta, omega and alpha.
 
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  • #10
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Another stab at part b:

x = x0 + v0t + (1/2)at^2
x = 0 + (.5 x 10) + .5(1)(10^2)
x = 5 + 50
x = 55m
 
  • #11
hage567
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Here's some pointers on your answers:

OK, so here's my crack at part a:

v = Vo + at
v = .5 + (1.0)(10)
v = .5 + 10
v = 10.5 m/s

Your answer looks good, but note that velocity is in m/s. Acceleration is m/s^2.

part b:

d = vt
d = (10.5)(10)
d = 105 m

You can't use d=vt, since this assumes a constant velocity (acceleration = 0). Since the ball is continuously accelerating at 1 m/s^2, you must find an equation that contains a as well.

part c:

OK, I think the question is asking how far the ball rolls during just the eighth second after being released from rest, not after 8 seconds from being released from rest. So your answer is not correct. For these calculations, now take Vo to be zero, which is the velocity at t=0 when the ball is released. I would start by finding the velocity at the beginning of the eighth second, and at the end of the eighth second.

v = Vo + at
v = .5 + (1.0)(8)
v = .5 + 8
v = 8.5 m/s

You can't take Vo=0.5 anymore. This is not the initial velocity at t=0. The initial velocity = 0, since the ball is being released from rest. Your answer is saying the velocity of the ball eight seconds after it reached the velocity of 0.5.

d = vt **see above for why you can't use this equation

d = vt
d = (8.5)(8)
d = 68 m

now was I correct in having to redo the velocity for part c?
 
  • #12
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I fixed the units, and I solved part b again.

a friend and I were wondering the same thing about whether it was AT 8 seconds or How Far it went from 8 to 9 seconds.
 
  • #13
hage567
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I fixed the units, and I solved part b again.

a friend and I were wondering the same thing about whether it was AT 8 seconds or How Far it went from 8 to 9 seconds.
Since it says during the eighth second, I would take it to be from t=7s to t=8s. If you count the interval between t=0s and t=1s as the first second. That's my opinion anyway.
 
  • #14
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Everything redone. Final Answer, Regis:wink:

part a:

v = Vo + at
v = .5 + (1.0)(10)
v = .5 + 10
v = 10.5 m/s

part b:

x = x0 + v0t + (1/2)at^2
x = 0 + (.5 x 10) + .5(1)(10^2)
x = 5 + 50
x = 55m

part c:

v = Vo + at
v = 0 + (1.0)(8)
v = 0 + 8
v = 8 m/s

x = x0 + v0t + (1/2)at^2
x = 0 + (8x1) + (1/2)(1.00m/s^2)(1s)^2
x = 8 + .5
x = 8.5m
 
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  • #15
hage567
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a) and b) look good. For c): If you are going to take the eighth second to be from t=8s to t=9s, then I agree with your answer is correct based on that.
 
  • #16
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a) and b) look good. For c): If you are going to take the eighth second to be from t=8s to t=9s, then I agree with your answer is correct based on that.
yeah, the problem is poorly worded if you ask me. I figure with the italics, it's stressing that point, so that's what I'm going with. I'll have to talk to my professor about it on Monday. Thanks for the help so far, as I have two more problems I have no idea about.
 

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