time and distance?

Question

A truck starts 1 foot from a motion detector and drives away at 2 feet per second for 6 seconds. If the speed were 3 feet per second instead of 2 feet per second and the starting point and number of seconds stayed the same, what would be the effect on the table of values? Oh, and the table of values is referring to that table of values-with time and distance

Answer 1

If the starting point is the same place for both trucks and the time over which the distance is measured is the same 6 seconds, the only difference is the final distance from the start point. The truck at 2 ft/sec over 6 sec = 12 ft traveled from the start point or 13 ft from the motion detector.

The truck at 3 ft/sec over 6 sec = 18 feet traveled from the start point or 19 feet from the motion detector.

This sounds like a trick question regarding the "motion detector".

Is there more information? What are you really asking?

Answer 2

Case 1. The distance d1 is given by d1 = 1 + 2t for 0 ≤ t ≤ 6.

Case 2. The distance d2 is given by d2 = 1 + 3t for 0 ≤ t ≤ 6.

Now simply calculate values for each of these for all integer t's from 0 through 6, and put them into a table. There will be three columns, the first headed " Time t ", the next two " Distance d1 (in ft.) " and " Distance d2 (in ft.) ".

For example, against the ' t ' value ' 4 ', you will have ' 9 ' in the ' d1 ' column and ' 13 ' in the ' d2 ' column.

If you merely wanted to summarise the effects on the time-distance table entries, you could just say that the ' d2 ' entries are greater then the ' d1 ' entries by ' 1 ' for every second elapsed.

Live long and prosper.

Answer 3

Consider this, for any rate, as time increases so does the distance; therefore, in order for distance to increase time or rate has to.