Set 1 Problem number 4

How long will it take an object to move 56 meters if it moves 7 meters every second?

Solution

Moving at 7 meters every second, we would go 7 meters in 1 second, 14 meters in 2 seconds, 21 meters in 3 seconds, . . ., 56 meters in 56 seconds. It helps to visualize this.

After visualizing it becomes clear that we can just divide the 56 meters by 7 meters / second to get the number of seconds required.
We also note that when we divide meters by meters/second we get seconds (m / (m/s) = m * (s / m) = (m * s) / m = s).

Generalized Solution

Generalized Response: Since displacement, or position change `ds is equal to the product of average velocity and time interval,

`ds = vAve `dt,

the time interval must be

`dt = `ds / vAve.

This is a simple algebraic rearrangement of `ds = vAve `dt (we just divide both sides by vAve), and it is also common sense: if we moved 6 meters at 2 meters per second we would require 3 seconds.

This is obvious if we think about it with obvious numbers (e.g., 20 meters at 5 m/s), and we observe that when the numbers are obvious we naturally divide the displacement by the time interval.
We do the same thing when the numbers aren't obvious.

Explanation in terms of Figure(s), Extension

Figure description:

The image below shows a 'triangle' connecting vAve, `dt and `ds. This triangle indicates that vAve,`dt and `ds are connected in such a way that the value of any of the three quantities can be found from the values of the other two.

We know that `ds = vAve `dt. If we divide both sides of this equation by `dt and reverse sides we get

vAve = `ds / `dt.

If we divide both sides by vAve and reverse sides we get

`dt = `ds / vAve.

These relationships are indicated on the diagram.

This specific problem could be solved by direct reasoning. We can also get the answer using the relationship `dt = `ds / vAve, with the given `ds and vAve.