# What is the speed of the ball 10.0 sec later?

• Peterson
In summary, the problem involves a ball rolling 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. The problems faced are to find the speed of the ball 10.0 sec later, the distance traveled in that 10.0 sec time interval, and the distance traveled during the eighth second after it was released. The solution involves using the basic kinematic equations of linear motion and plugging in the known values to solve for the unknowns. After some corrections, the final answers are 10.5 m/s for part a, 55 m for part b, and 8 m/s and 8 m for
Peterson
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). I am pretty much at a loss here. Any help for a lazy, "I just want to graduate" senior would be of great help.

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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.

hage567 said:
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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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 traveling 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 formulay do you think you could use?

ok, you answered my biggest problem. 0.5 is the starting point.

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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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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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

Peterson said:
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?

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.

Peterson said:
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.

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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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.

hage567 said:
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.

## 1. How is the speed of a ball defined?

The speed of a ball is defined as the rate at which the ball covers distance over a period of time. It is a measure of how fast an object is moving.

## 2. Why is it important to measure the speed of a ball?

Measuring the speed of a ball can help scientists understand the motion and behavior of different objects. It is also useful for predicting the trajectory of a ball and determining its impact force.

## 3. What factors affect the speed of a ball?

The speed of a ball can be affected by various factors such as the initial velocity, air resistance, gravity, and surface friction. These factors can either increase or decrease the speed of the ball.

## 4. How is the speed of a ball calculated?

The speed of a ball can be calculated by dividing the distance the ball travels by the time it takes to travel that distance. This is known as the average speed formula: speed = distance/time. It can also be calculated by using the equation for velocity: velocity = change in distance/change in time.

## 5. Can the speed of a ball change over time?

Yes, the speed of a ball can change over time. This can be due to external forces acting on the ball, such as wind or friction. The speed can also change if the ball is undergoing acceleration or deceleration.

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