Salem and vernonville are 200 miles apart. A car leaves Salem traveling towards vernonville, and another car leaves vernonville at the same time, traveling towards Salem. The car leaving Salem averages 10 miles per hour more than the other, and they meet after 1 hour 36 minutes. What are the average speeds of the cars?

car leaving salem = 65 mph care leaving the other place- 55 mph

Quote:

---

Originally Posted by Heb0717

car leaving salem = 65 mph care leaving the other place- 55 mph

---

First - it's against our policy on this board to simply give homework answers away. And second - your answer is incorrect.

To the OP: please show us your attempt at this and we'll help where you get stuck.

I'll give you some hints to solve the problem

For this kind of problem, there are 3 things we need:
Distance, speed and time. You want to find speed/average velocity. So we need Distance and Time for it

Assume:

For distance
x is distance Salem driver travel and until he meets the other driver
Therefore, 200-x should be the distance the other driver travels to meet Salem's driver

For time:
They both leave at the same time, and meet at the same time, so there travel time should be the same which is already given 1h 36 minutes. But you need to convert 1h 36mins into hour only. Call time here is "t"

So speed formula, velocity = distance / time

So for the speed of first driver from Salem:
V1 = x/t
Speed of second driver:
V2 = (200 - x)/t

The last hint in the problem is the first driver travel 10 mph faster than the other one.
So

V1 - V2 = 10

Or
(x/t) - (200-x)/t = 10

*Remember: time in here is already given to you
So the only last unknown is x needed to be solve. When you have x, plug back into V1 and V2 to solve for average speed of each driver.