Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. How do you determine speed of the vehicle?

Let `s` be the speed (in miles/hour) that Steve was traveling for `h` hours to cover `200` miles.

We are told that if he had traveled at a speed of `(s+10)` miles/hour that it would have taken him `(h-1)` hours to cover the `200` miles.

Since distance travelled `=`speed `xx` time

`color(white)("XXX")200 = sh`
`color(white)("XXXXXXXXXXX")rarr color(blue)(h)=color(green)(200/s)`
and
`color(white)("XXX")200=(s+10)(color(blue)(h)-1))`

So we have
`color(white)("XXX")200=(s+10)(200/s-1)`

`color(white)("XXXXXX")=s(200/s-1)+10(200/s-1)`

`color(white)("XXXXXX")=200-s+2000/s-10`

`rArrcolor(white)("XXX")s-2000/s+10=0`

`rarrcolor(white)("XXX")s^2+10s-2000=0`

Using the quadratic formula
`color(white)("XXX")s=(-10+-sqrt(10^2-4(1)(-2000)))/(2(1))`

`color(white)("XXXX")=(-10+-sqrt(8100))/2`

`color(white)("XXXX")=(-10+-90)/2`

`s=-25` or `s=40`

Since the speed must be non-negative, `s=-25` is an extraneous solution.

There are 2 different scenarios described here, Write an expression for the speed of each of them.

The difference between the times would be 1 hour. This allows us to make an equation.

Let the slower speed be `x` miles per hour.
The faster speed is `x +10` miles per hour.

`time ="distance"/"speed"`

At the slower speed, the time, `T_1 = 200/x` hours

At the faster speed, the time, `T_2 = 200/(x+10)`hours

(`T_1 "will be more than " T_2` because if we drive at slower speed, the journey will take longer.)
The difference between the two times is 1 hour.

`T_1 - T_2 = 1`

`200/x -200/(x+10) = 1 " now solve the equation"`

Multiply each term by `color(red)(x(x+10))`

`(color(red)(x(x+10))xx200)/x -(color(red)(x(x+10))xx200)/(x+10) = color(red)(x(x+10))xx1`

`(cancelx(x+10)xx200)/cancelx -(xcancel((x+10))xx200)/cancel((x+10)) = x(x+10)xx1`

`200x+2000 -200x =x^2+10x`

`x^2 +10x -2000 = 0`

Find factors of 2000 which differ by 10.

The factors must be quite close to `sqrt2000`, because there is a very small difference between them.

We find `40xx50 = 2000`

`(x-40)(x+50) = 0`

`x = 40 or x = -50,` (reject -50)

The slower speed is 40mph, the faster speed is 50 mph.