Set 2 Problem number 2

The velocity of an object increases at a constant rate of 6 meters/second per second.

How long will it take the object to increase its velocity by 10 meters/second?

Solution

The second-by-second increases are 6, 12 , 18, ... meters per second.

At some point the increase will reach 10 meters per second.
The exact time is ( 10 meters/sec) /( 6 meters/sec/sec) = 1.666667 sec.

In the language of rates, velocity is changing at the rate of 6 (m/s) / s, or 6 m/s² (note that (m/s) / s = (m/s) * (1/s) = m/s^2).

Changing at the rate of 6 (m/s) / s, it will require

`dt = 10 m/s / ( 6 (m/s) / s) = 1.666667 sec for the change to occur.

We have divided the change in the quantity (velocity) by the rate (acceleration) to get the time interval.

Generalized Response: If we let a stand for the number of meters per second the velocity increases in a second, and `dt the number of seconds, then the velocity increase in `dt seconds will be the product a `dt of the velocity increase per second and the number of seconds.

If we represent velocity increase by `dv, we have

`dv = a `dt.

We understand a as the rate at which velocity increases. We call a the acceleration of the object.

This statement was in some ways clearer when we were thinking in terms of meters per second per minute, as on the last problem group.
In this example the seconds in meters per second get mixed up with the seconds over which velocity increases. So it is important to understand acceleration in terms of meters per second per minute before thinking of it in meters per second per second. However the units used in this problem are standard units, and we will generally measure acceleration in meters/second/second.

If we understand a as the rate at which velocity changes, then we see that the change in velocity is the product of the rate a and the time interval `dt:

`dv = a `dt.

From this we easily see that `dt = `dv / a.

Explanation in terms of Figure(s), Extension

The first figure below shows how the average rate aAve of velocity change and the time interval `dt give us the amount `dv of the velocity change.

We would for example multiply the velocity change per minute by the number of minutes to get the net change. In general we find `dv by multiplying a by `dt:

`dv = aAve `dt.

The second figure completes the first by indicating how any of the quantities aAve, `dt and `dv are related.

For the present problem we would use the form `dt = `dv / `aAve.