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Among other things this problem should illuminate the differences among

• midpoint clock time,

• change in clock time,

• clock time,

• time interval,

• initial velocity,

• final velocity,

• average velocity,

• change in velocity

• the meaning of the slope of a position vs. clock time graph and

• the meaning of the slope of a position vs. clock time graph.

A ball starts with velocity 0 and accelerates uniformly down a ramp of length 30 cm, covering the distance in 5 seconds.

• What is its average velocity? 

Its average velocity is the average rate at which position changes with respect to clock time, which in turn is equal to (change in position) / (change in clock time), so we have

• vAVe = (30 cm) / (5 sec) = 6 cm / sec.

• If the acceleration of the ball is uniform then its average velocity is equal to the average of its initial and final velocities. 

You know its average velocity, and you know the initial velocity is zero. 

What therefore must be the final velocity? 

The initial velocity of the ball is 0 and acceleration is uniform.  Therefore its final velocity is

• vf = 2 * vAVe = 2 * (6 cm / sec) = 12 cm/s.

• By how much did its velocity therefore change? 

The velocity of the ball changed from its initial velocity v0 = 0 cm/s to its final velocity vf = 12 cm/s.  So the change in velocity is

• `dv = vf - v0 = 12 cm/s - 0 cm/s = 12 cm/s.

• At what average rate did its velocity change with respect to clock time? 

The average rate of change of velocity with respect to clock time is by definition

• ave rate =

(change in velocity) / (change in clock time) =

12 cm/s / (5 s) =

2.4 cm/s^2.

• What would a graph of its velocity vs. clock time look like?  Give the best description you can.

The reasoning here is as follows:

• Acceleration is uniform, meaning it is unchanging. 

• Acceleration is the rate of change of velocity with respect to clock time, which is represented by the slope of the v vs. t graph. 

• The slope of the graph is therefore constant. 

• The graph of v vs. t rises along a straight line from v = 0 to v = 30 cm/s while clock time t changes by 5 seconds. 

• The rise of the graph between these points is 30 cm/s, the run is 5 s, and the slope of the line is rise / run = (30 cm/s) / (5 s) = 6 cm/s^2. 

• The slope is identical to the (uniform) acceleration. 

Additional Suggestions

Any uniformly accelerated motion problem can be reasoned out from the three things listed below. On every problem you should start by writing down all three, and you should continue writing down these definitions, and applying them very carefully, until you are very sure of them. 

• the definition of average velocity

• the definition of average acceleration.

• a short note that if acceleration is uniform, then average velocity is the average of the initial and final velocities

To answer the question 'by how much does velocity change' on an interval you should ask yourself the following three questions:

• What was the velocity at the beginning of the interval?

• What was the velocity at the end of the interval?

• What therefore was the change in velocity?

To apply the definition of the average rate of change of velocity with respect to clock time:

• What is this definition?

• What is the change in velocity?

• What is the change in clock time?
   

It is important to distinguish the graph of velocity vs. clock time from the graph of position vs. clock time. 

• If acceleration is uniform, the graph of v vs. t is a straight line. The slope of this line is the acceleration.

• The graph of position vs. t has slopes that represent velocities. If acceleration is constant (and not zero) velocities are continually changing, and the slopes of this graph change accordingly, giving the graph a curved shape.

Don't confuse average velocity with change in velocity or with average rate of change of velocity with respect to clock time.

• Average velocity is change in position / change in clock time. 

• If the v vs. t graph on an interval is a straight line segment, the average velocity on that interval occurs at the midpoint of the segment.

• If the v vs. t graph on an interval is a straight line, then the average velocity is equal to the average of the initial and final velocities,

vAve = (final velocity + initial velocity) / 2.

• The change in velocity on an interval is

`dv = (final velocity) - (initial velocity). 

This quantity is unrelated to the average velocity. 

• The average rate of change of velocity with respect to clock time on an interval follows from the definition of rate of change, and is given by

ave rate = (final velocity - initial velocity) / (change in clock time)

This quantity is be definition equal to the average acceleration.

• In summary, don't confuse the three quantities average velocity, change in velocity and average acceleration.  Know which is which and know which applies to the question you are answering or the situation you are analyzing:

• vAve = `ds / `dt, which also is equal to (vf + v0) / 2 if acceleration is uniform

• `dv = vf - v0

• aAve = (vf - v0) / `dt