Set 2 Problem number 5

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

An object is observed for 7 seconds, during which time its velocity is seen to increase steadily from 5 meters per second to 13 meters per second.

• On the average, by how many meters per second does velocity increase, per second, while being observed?

Solution

Reasoning intuitively we see that velocity increases by ( 13 - 5) meters per second = 8 meters per second in 7 seconds.

• Per second, this is an increase of 8/ 7 meters per second, or 1.142857 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 8 m/s in 7 seconds.
• This implies an average rate of 8 m/s / ( 7 s) = 1.142857 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)