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course Phy 201
Problem 1#$&*
course Phy 201
If the velocity of the object changes from 3 cm / sec to 15 cm / sec in 10 seconds, then at what average rate is the velocity changing?
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The average rate equation is defined as the change in a divided by the change in b, here a is the change in velocity and b is the change in time, which is acceleration. The velocity is changing at an average rate of:
=(15cm/s - 3cm/s)/10s
=1.2cm/s^2
Velocity with respect to time represents the acceleration.
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You are using the right relationship, and the right quantities are in the right places.
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12 cm/s / (10 s) has a numerical value of 1.2, but (cm/s) / s does not give you cm, nor is this a unit appropriate to rate of change of velocity with respect to clock time.
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A ball rolling from rest down a constant incline requires 4.4 seconds to roll the 86 centimeter length of the incline.
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What is its average velocity?
vAve=delta s/ delta t
=86cm/4.4s
=19.54cm/s
An object which accelerates uniformly from rest will attain a final velocity which is double its average velocity.
What therefore is the final velocity of this ball?
The average velocity of a uniformly accelerating object that starts from rest (only) will have an average velocity that is half it’s final velocity. Doubling the average velocity, 19.54cm/s, gives 39.09m/s, the final velocity.
The velocity change:
39.09cm/s - 19.54cm/s = 19.54cm/s
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This is not the change in velocity form the beginning of the 4.4 s interval to the end.
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The clock time change:
tf - ti= 4.4s - 0s = 4.4s
Therefore, the requested average rate is the change in velocity divided by the change in time, the acceleration:
aAve= delta v/ delta t
aAve= 19.54cm/s / 4.4s
aAve=4.44cm/s^2
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You're on the right track.
How much did the velocity change during the entire interval specified in the problem?
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This is the one question you haven't answered (at least not correctly), and of course it has a bearing on the requested rate.
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By how much did the clock time change?
What therefore is the requested average rate?
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Do be careful about the units of your calculation.
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An automobile accelerates uniformly down a constant incline, starting from rest. It requires 10 seconds to cover a distance of 138 meters. At what average rate is the velocity of the automobile therefore changing?
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To find the average rate the velocity is changing with respect to time, we need to find the average acceleration, but first, we must calculate the average velocity to solve for average acceleration:
vAve=delta s/ delta t
vAve=138m/10s
vAve=13.8m/s
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13.8 m/s is the average velocity on the interval, but it is not the change in velocity.
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aAve=delta v/ delta t
aAve=13.8m/s / 10s
aAve=1.38m/s^2
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This is the average rate of change of position with respect to clock time.
This is useful, but you were asked for the average rate of change of velocity with respect to clock time.
The reasoning of the preceding questions are relevant to the solution of this problem as well.
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You are following the procedure quite well, but you are not getting the correct changes in velocity.
Please use #### for insertions on the next revision.
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Be sure to include the entire document, including my notes. &#
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