Question

Ahmad and George took a circular route of 7 km that starts and ends at the same point?

Ahmad and George took a circular route of 7 km that starts and ends at the same point. They started at the same time, took the route in opposite directions, and finished at the same time. Ahmad walked at a constant speed all along the route. George walked at a constant speed for the rst 3 km then increased his speed by 7 km/hr, and kept this constant speed for the remaining distance. They met only once during the walk, that was, when Ahmad had covered 4:5 km of the distance. How many hours did it take them to finish the 7 km distance?

Answer 1

`t = 1.94` `"h"`

Explanation:

WARNING: LONG ANSWER (possibly a little confusing at times...)

This question is worded slightly odd, but I'll do my best in the assumptions. We're asked to find how long it takes the two of them to complete a `7`-`"km"` circular path.

We can model this question in a slightly different way; instead of their motion being in a circle, let's make it a straight line, i.e. the `x`-axis with Ahmad beginning at `x = 0` and George beginning at `x = 7` `"km"` (they're traveling in opposite directions).

When you said "Ahmad had covered 4:5 km of the distance" I'll assume you mean he covered `4.5` `"km"` when they met.

If they met at `x = 4.5` `"km"`, that means George covered a distance of `7` `"km"` `- 4.5` `"km"` `= 2.5` `"km"` when they met. We know that George didn't change his speed until he traveled `3` `"km"`, so when they met, neither of them had previously changed their speed.

We'll call Ahmad's speed the variable `v`, and George's speed `v_G`. Their equations for their speeds are

• Ahmad: `v = (4.5color(white)(l)"km")/t`

• George: `v_G = (2.5color(white)(l)"km")/t`

We know the time `t` is the same when they traveled these distances, because this is the instant their positions were the same. We can therefore rearrange each equation to solve for `t`, and set them equal to each other:

• Ahmad: `t = (4.5color(white)(l)"km")/v`

• George: `t = (2.5color(white)(l)"km")/(v_G)`

Therefore,

`(4.5color(white)(l)"km")/v = (2.5color(white)(l)"km")/(v_G)`

From this, George's speed in terms of Ahmad's (`v`) is

`v_G = 5/9v`

George travels this speed for `3` `"km"`, then for the next `4` `"km"` he travels with a speed of

`5/9v + 7` `"km/h"`. (he increased his speed by `7` `"km/h"`)

Now what we can do is find the complete descriptions for each person's motion, in terms of the time, `t`.

For Ahmad, we have

`t = (7color(white)(l)"km")/v`

And for George:

`t = overbrace((3color(white)(l)"km")/(5/9v))^"3 kilometers with this speed" + overbrace((4color(white)(l)"km")/(5/9v + 7color(white)(l)"km/h"))^"4 kilometers with this speed"`

Let' sset these two equations equal to each other again, and solve for the speed, `v`:

`(7color(white)(l)"km")/v = (3color(white)(l)"km")/(5/9v) + (4color(white)(l)"km")/(5/9v + 7color(white)(l)"km/h")`

If you don't want to know how to solve this, skip the indented portion.

To simplify things, we'll neglect units in our work:

`7/v = 3/(5/9v) + 4/(5/9v + 7)`

`7/v = 27/(5v) + 4/(5/9v + 7)`

Note that the quantity

`4/(5/9v + 7)` can also be written as

`36/(5v + 63)` (multiplied numerator and denominator by `9`)

So we then have

`7/v = 27/(5v) + 36/(5v + 63)`

The two quantities on the right side can be simplified by a common denominator `5v(5v + 63)`:

`7/v = (63(5v + 27))/(5v(5v+ 23))`

Cross multiply:

`35v(5v+63) = 63v(5v + 23)`

`175v^2 + 2205v = 315v^2 + 1701v`

`140v^2 = 504v`

`140v = 504`

`v = color(red)(3.6` `color(red)("km/h"`

Now that we know Ahmad's speed `v`, let's use this to find the time it takes to complete one revolution:

`t = (7.0cancel("km"))/(color(red)(3.6)(cancel(color(red)("km")))/(color(red)("h"))) = color(blue)(1.94` `color(blue)("h"`