Question

Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards​ Vernonville, and another car leaves Vernonville at the same​ time, traveling towards Lewiston. The car leaving Lewiston averages 10 miles per hour more than the​ other, and they meet after 1 hour and 36 minutes. What are the average speeds of the​ cars?

Answer 1

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Step-by-step explanation:

Average speed of the car A is
`v_(A) =(x_(A) )/(t_(A) )` (Equation A) and Average speed of the car B is
`v_(B) =(x_(B) )/(t_(B) )` (Equation B), where
`x_(A)` and
`x_(B)` are the distances and
`t_(A)` and
`t_(B)` are the times at which are travelling the cars A and B respectively.

We have to convert the time to the correct units:

1 hour and 36 minutes = 96 minutes

`96 minutes . (1 hour)/(60 minutes) = 1.6 h`

From the diagram (Please see the attachment), we can see that at the time they meet, we have:

`v_(A) = (208-x)/(1.6h) + 10(miles)/(h)` (Equation C)

`v_(B) = (208-x)/(1.6h)` (Equation D)

From Equation A and C, we have:

`(208-x)/(1.6)+10 = (x)/(1.6)`

208-x+16 = x

208 + 16 = 2x

`x = (224)/(2)`

x = 112 miles

Replacing x in Equation A:

`v_(A)  = (112miles)/(1.6h)`

`v_(A) = 70 miles per hour`

Replacing x in Equation B:

`v_(B)  = (208miles-112miles)/(1.6h)`

`v_(B)  = (96miles)/(1.6h)`

`v_(B)  = 60 miles per hour`