Set 2 Problem number 5

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object is moving at a constant rate before, and at another constant rate after, a constant-acceleration phase lasting 8 seconds.

• What is the rate at which velocity changes during the acceleration phase, in m/s per second?

Solution

Reasoning intuitively we see that velocity increases by ( 11 - 7) meters per second = 4 meters per second in 8 seconds.

• Per second, this is an increase of 4/ 8 meters per second, or .5 meters per second per second.

More formally, we are looking for the average rate of velocity change, which is the acceleration.

• We have a velocity change of 4 m/s in 8 seconds.
• This implies an average rate of 4 m/s / ( 8 s) = .5 m/s²
• Since velocity changes at a constant rate the average rate can simply be called the rate.

Generalized Solution

The change in velocity is the difference `dv = vf - v0 between the initial and final velocities.

• The rate at which velocity changes is the velocity difference divided by the time interval:

• rate of velocity change = `dv / `dt = (vf - v0) / `dt.

Explanation in terms of Figure(s), Extension

The figure below shows (blue lines) how the rate a at which velocity increases is obtained from `dv and `dt.

• This relationship is part of the triangle relating a, `dv and `dt.
• This relationship tells us that a = `dv / `dt.

The figure also shows how `dv is obtained in the obvious way from v0 and vf:

• `dv is just the difference vf - v0 between the velocities.

Note that the full relationship a = `dv / `dt = (vf - v0) / dt is given.

• You should understand what this relationship tells you:

• Acceleration is the rate `dv / `dt at which velocity changes, and since `dv = vf - v0, acceleration is a = (vf - v0) / `dt.

Figure(s)