Question

I cannot get this right. The fraction (total time) is messing me up. Can you provide a step-by-step for this one?

Answer 1

He drives the first leg at an average of 24 mph (taking 1.25 h), and the return trip at 20 mph (taking 1.5 h)

Explanation:

We use

`d=vt`

to express this distance, and `v_1` and `t_1` for the first trip (to Burlington) and `v_2` and `t_2` for the return.
Then

`v_1t_1 =30` and

`v_2t_2=30`

From the problem, we know that `v_2=v_1-4` and that `t_1+t_2=2.75` hr.

Using this in the second equation

`(v_1-4)(2.75-t_1)=30`

Expanding this, we get

`-v_1t_1+4t_1+2.75v_1-11=30`

But we know `v_1t_1=30`, so the equation can be written

`-30+4t_1+2.75v_1-11=30`

`2.75v_1+4t_1 - 71=0`

and since `t_1 = 30/(v_1)`

`2.75v_1+4(30/v_1) - 71=0`

or

`2.75(v_1)^2-71v_1+120=0`

We solve with the quadratic formula

`v_1=(71+-sqrt((-71)^2-4(2.75)(+120)))/(2(2.75))`

`v_1=24` mph

Which means `v_2=20` mph

(1.25 h on the trip out, 1.5 h on the return)

Answer 2

Patrick travels at an average speed of `"24 mi/h"` during his trip and `"20 mi/h"` when returning.

Explanation:

We know that `r= d/t`, where `r` is the rate (speed), `d` is the distance, and `t` is the time.

In this case, `d = "30 mi"`.

Let `r_1` be the speed of the first trip and `r_2` be the speed of the return trip. We are told that the speed for the return trip is `"4 mi/h"` slower, so `color(blue)(r_2 = r_1 - 4)`.

Since the total time of the round trip is `2 3/4` or `"2.75 h"`, we can say `color(green)(t_1 + t_2 = 2.75)`.

We have two equations modeling the speeds:

`r_1 = 30 / t_1` and `r_2 = 30 / t_2`

Let's get the variables to match in both equations.

`t_1 = 30/r_1` and `t_2 = 30/r_2`

`color(green)(t_1 + t_2) = 30/r_1 + 30/r_2`

`color(green)2.75 = 30/r_1 + 30/color(blue)(r_2)`

`2.75 = 30/r_1 + 30/(color(blue)(r_1 - 4))`

Now, solve for `r_1`.

`2.75 - 30/r_1 = 30/(r_1 - 4)`

`(2.75r_1) / r_1 - 30/r_1 = 30/(r_1 - 4)`

`(2.75r_1 - 30)/r_1 = 30/(r_1 - 4)`

`(2.75r_1 - 30)(r_1 -4) = 30r_1`

`2.75r_1^(color(white)x2) - 11r_1 - 30r_1 + 120 = 30r_1`

`2.75r_1^(color(white)x2) - 71r_1 + 120 = 0`

We can use the Quadratic Formula to find `r_1`.

`r_1 = (-b +- sqrt(b^2 - 4ac))/(2a)`

`r_1 = (71 +- sqrt ((-71)^2 - 4(2.75)(120))) / (2(2.75))`

`r_1 = 24` or `1.bar81`

We can throw out `r_1 = 1.bar81` because we know that `r_2` must be `"4 mi/h"` slower. Thus, `r_1 = "24 mi/h"` and `r_2 = "20 mi/h"`.