cq1_08_1 solution and discussion

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A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.

The acceleration of gravity and the initial velocity of the ball are in opposite directions. One is upward and the other is downward. So they have different signs.

We need to choose either upward or downward as the positive direction. We could choose either, but having chosen the positive direction all of our displacements, velocities and accelerations will be either positive or negative, depending on whether they are in the chosen positive direction or opposite to it.

In the following solution we will choose the upward direction as positive. This means that the downward direction is negative.

What will be the velocity of the ball after one second?

The ball starts out at 25 m/s and changes velocity by -10 m/s every second. So after 1 second it is traveling at 15 m/s, and after 2 seconds at 5 m/s.

What will be its velocity at the end of two seconds?

Its average velocity for the first second is the average of its 25 m/s initial velocity and its 15 m/s final velocity on that interval. So average velocity is 20 m/s and the displacement is `ds = vAve `dt = 20 m/s * 1 s = 20 m.

During the first two seconds, what therefore is its average velocity?

How far does it therefore rise in the first two seconds?

Its average velocity for the first 2 seconds is the average 15 m/s of its initial 25 m/s velocity and its final 5 m/s velocity on this interval, and its displacement is `ds = vAve * `dt = 15 m/s * 2 = 30 m.

What will be its velocity at the end of a additional second, and at the end of one more additional second?

After 4 seconds it is moving at -15 m/s and its average velocity for the 4-second interval is 5 m/s, the average of 25 m/s and -15 m/s. Its displacement is 5 m/s * 4 s = 20 m.

At what instant does the ball reach its maximum height, and how high has it risen by that instant?

What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?

It reaches its maximum height at the instant it comes to rest. To come to rest from 25 m/s, accelerating at -10 m/s^2, requires 2.5 seconds. The average velocity on this interval is 12.5 m/s, the average of its 25 m/s initial velocity and its 0 m/s final velocity on this interval. At an average of 12.5 m/s the ball will travel 12.5 m/s * 2.5 s = 31.25 meters, and this is the maximum height to which it will rise.

How high will it be at the end of the sixth second?

If acceleration is uniform for the entire six seconds, the ball's final velocity will be -35 m/s, its average velocity will be (25 m/s + (-35 m/s) ) / 2 = -5 m/s, and its height will be -5 m/s * 6 s = -30 m relative to its starting point.

This same result would be obtained from the equations of motion using v0 = 25 m/s, a = -10 m/s^2 and `dt = 6 s.

Note that there is no need to find the maximum height, or to solve a motion problem which starts at the maximum height.

Of course if the ground gets in the way before the sixth second, the acceleration ceases to be uniform and the uniform-acceleration analysis stops at that instant.

Neither the position of the ground, or the initial position of the ball relative to the ground, are mentioned in this problem. Since the ball might have been thrown from 30 m or higher, this solution could make sense.

The correct solution would assume uniform acceleration for the entire 6 seconds, then note that if something interferes with the uniformity of the acceleration, the solution wont necessarily correspond to reality. Its also possible that the ball was thrown from the ground, but that there's a hole into which the ball could fall.

The velocity after 6 sec is -35 m/s so ave vel for the six sec is -5 m/s and displacement is -30 m. If the ball is thrown from a point 30 m high or more, this solution might be valid. At those speeds air resistance is likely to be significant so the solution won't be completely accurate.

EXERCISE FOR THE PERPLEXED

If you are having trouble with this problem and think it would be helpful, you are welcome to submit this part of the document:

There are a number of questions here, but the answers are short so this won't necessarily take you very long. If it takes you more than 30 minutes, submit what you have so I can respond with help and suggestions. Use both the form and email. Simply submit a copy of the entire document, with your answers to these questions (and/or questions you might have) inserted in the appropriate places:

The definition of average velocity is average rate of change of position with respect to clock time, which is equal to

ave velocity = (change in position) / (change in clock time).

In symbols we would write this

vAve = `ds / `dt.

Suppose now that something travels at 3 cm/s for 5 seconds.

Is 3 cm/s the ave velocity, the change in position, or the change in clock time?
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Is 5 seconds the ave velocity, the change in position, or the change in clock time?
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If you know the change in position and the change in clock time, how do you find the average velocity?
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If you know the change in position and the average velocity, how do you find the change in clock time?
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If you know the change in clock time and the average velocity, how do you find the change in position?
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Is 3 cm/s the value of vAve, `ds or `dt?
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Is 5 s the value of vAve, `ds or `dt?
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If you know `ds and `dt, how do you find vAve?
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If you know vAve and `dt, how do you find `ds ?
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If you know `ds and vAve, how do you find `dt?

The definition of average acceleration is average rate of change of velocity with respect to clock time, which is equal to

ave acceleration = (change in velocity) / (change in clock time).

In symbols we would write this

aAve = `dv / `dt.

Suppose now that something something accelerates at 7 cm/s^2 for 3 seconds.

Is 7 cm/s^2 the ave acceleration, the change in position, or the change in clock time?
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Is 3 seconds the ave acceleration, the change in position, or the change in clock time?
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If you know the change in velocity and the change in clock time, how do you find the average acceleration?
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If you know the change in velocity and the average acceleration, how do you find the change in clock time?
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If you know the change in clock time and the average acceleration, how do you find the change in velocity?
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Is 7 cm/s the value of aAve, `dv or `dt?
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Is 3 s the value of aAve, `dv or `dt?
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If you know `dv and `dt, how do you find aAve?
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If you know aAve and `dt, how do you find `dv?
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If you know `dv and aAve, how do you find `dt?
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