Question

To train for the running portion of the race,she runs 7 miles each day over the same course. The first 3 miles of the course is on level ground, while the last 4 miles is downhill. She runs 2 miles per hour slower on level ground than she runs downhill?

If the complete course takes 1 hour, how fast does she run on the downhill part of the course?

Answer 1

8 mph

Explanation:

So we know that the total run will take 1 hour. 3 miles will be at `x-2` mph and 4 miles will be at `x` mph.

Therefore, the amount of time she spends doing the first 3 miles is `3 / (x-2)` hours and she spends `4 / x` hours running the rest. Since that adds up to 1 hour,

`3/(x-2) + 4/x = 1`
`(3x + 4(x-2))/(x(x-2)) = 1`
`3x + 4x-8 = x(x-2)`
`0 = x^2 - 9x +8`
By quadratic formula,
`x = (9 pm sqrt(81 - 4 cdot 8))/(2) = (9 pm 7)/2`
Since we want the faster speed, `x = 16 / 2 = 8`.

Therefore, she runs on flat ground at 6 miles per hour and downhill at 8 miles per hour.

We can check this: it would take her 30 minutes to run 3 miles at 6 miles per hour and it would take her 30 minutes to run 4 miles at 8 miles per hour, so it would take her an hour to run all 7 miles, as we wanted.

Answer 2

A LOT OF DETAIL is given so you can see where everything comes from. Building simultaneous equations.

5 miles per hour

Explanation:

Units of measurement in green
Values in red

Let the 'velocity' on level ground be `x`
Let the time run for level ground be `t_L`
Let the time run for downhill be `t_d`

For level ground `(color(red)(x)color(white)(.)color(green)(m/cancel(h)^1))xx(color(red)(t_L)color(white)(.)color(green)(cancel(h)^1))=color(red)(3)color(white)(.)color(green)( m)color(white)(.) Equation(1)`

For down hill `(color(red)(2x)color(white)(.) color(green)(m/cancel(h)^1)) xx (color(red)(t_d)color(white)(.)color(green)(cancel(h)^1)) = color(red)(4)color(white)(.)color(green)( m)color(white)(.) .......Equation(2)`

Total time running `->1 hour -> color(red)(t_L+t_d=1)color(white)(.)color(green)( h)" "..Equation(3)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the value of "x)`

Consider `Equation(3)`

Subtract `t_d` from both sides.

`color(red)(t_L=1-t_d)" "..................Equation(3_a)`

Using `Equation(3_a)` substitute for `t_L` in `Equation(1)`
This changes the problem into simultaneous equations.

`x(1-t_d)=3" ".........................Equation(1_a)`
`2xt_d=4" "..................................Equation(2)`

`-xt_d+x=3" ".........................Equation(1_a)`
`color(white)(.)2xt_dcolor(white)("ddd")=4" "...........................Equation(2)`

Multiply `Eqn(1_a)` by 2

`-2xt_d+2x=6" ".........................Equation(1_a)`
`ul(color(white)(..)2xt_dcolor(white)("dddd")=4" "...........................Equation(2))`
`color(white)("ddddddd")2x=10 larrEqn(1_a)+Eqn(2)`

Divide both sides by 2

`color(blue)(x=5)`

`color(blue)("=========================================")`

`color(blue)("Determine the time running on the level")`

Let velocity (spead or rate) be represented by `v`

Substitute for `x=5` in `Equation(1)` to determine `t_L`

For level ground `(color(red)(x)color(white)(.)color(green)(m/cancel(h)^1))xx(color(red)(t_L)color(white)(.)color(green)(cancel(h)^1))=color(red)(3)color(white)(.)color(green)( m)color(white)(.) Equation(1)`

`5t_L=3 => t_L=3/5 " hours"`

Thus velocity (rate) `->"velocity"xx"time"="distance"`

`color(white)("dddddddddddddddddddd")vcolor(white)("dd.")xxcolor(white)("dd")t_L=3`

`v=3-:t_L`

`v=3-:3/5`

`v=3xx5/3 = 5 mph`