cq_1_021

Your 'cq_1_02.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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Here is a narrative solution by which you can check yourself. Try to answer the questions I've posed above without reference to this solution, then check it out to see how you did:

Solution:

A ball starts with velocity 4 cm/sec and ends with a velocity of 10 cm/sec.

What is your best guess about the ball's average velocity?

The lesser of the two known velocities is the 4 m/s initial velocity. The greater of the two is the 10 m/s final velocity. Lacking other information, the most reasonable estimate of the average velocity would be the velocity halfway between these two, which is easily found by averaging the two velocities to get

(4 cm/s + 10 cm/s) / 2 = 7 cm/s.

Without further information, why is this just a guess?

It is certainly possible that the ball changed velocity at a nonuniform rate. The ball might even have slowed down below 4 cm/s before speeding up to 10 cm/s, or it might have exceeded 10 cm/s at some point during the interval; it might have spent most of the interval moving at 4 cm/s before suddenly speeding up to 10 cm/s. In any of these cases, the velocity does not change at a uniform rate, and our 7 cm/s estimate of the average velocity could differ significantly from the average velocity.

If it takes 3 seconds to get from the first velocity to the second, then what is your best guess about how far it traveled during that time?

Recall that the average velocity is the average rate of change of position with respect to clock time so that

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

The change in position is therefore

change in position = ave vel * change in clock time

(you can find this algebraically by just multiplying the both sides of the equation for ave vel by change in clock time).

Using the average velocity 7 cm/s, we therefore find that

change in position = 7 cm/s * 3 s = 21 (cm/s) (s) = 21 cm.

At what average rate did its velocity change with respect to clock time during this interval?

The average rate of change of velocity with respect to clock time is (change in velocity) / (change in clock time). The change in velocity from 4 cm/s to 10 cm/s is 10 cm/s - 4 cm/s = 6 cm/s. If this occurs in 3 seconds, then we have

average rate of change of velocity with respect to clock time = (change in velocity) / (change in clock time) = 6 cm/s / (2 s) = 3 cm/s^2.