Average rate son was traveling

Question

Ashley left Weston on her bicycle riding at an average rate of 10 miles per hour three hours before her son left by automobile.The son overtook Ashley in exactly 30 minutes. At what average rate was Ashley's son traveling.

Answer 1

Hi Jay,

 

To solve for the average rate of the automobile, we need the equation that relates speed, distance, and time. That equation is speed = distance / time.

 

Looking at the first statement, what equation can we make with the knowledge that Ashley was traveling at an average of 10 mph for 3 hours? Is it possible to determine the distance she traveled during this period of time? How far did her son travel in this period?

 

Now looking at the second statement, is it possible to determine the distance Ashley traveled during the 30 minute period? What equation can we set up for the distance that her son traveled in the 30 minutes?

 

The key to solving this problem is to use the equation for speed in each traveling period and to determine the distance that Ashley traveled in those periods and the distance her son traveled in those periods. What do we know about Ashley's distance compared to her son's distance at the point where the son overtakes Ashley?

 

Let me know if you have any other questions about this problem. 

Answer 2

Since Ashley left 3 hours before her son and was averaging a speed of 10 mph, she had a 30 mile head start on her son.  

 

Let x = son's speed

 

Distance = (Rate)(Time)

 

Ashley's son catches up to her when their distances are equal

 

Son's distance in 30 minutes (1/2 hour) = (1/2)(x)

 

Ashley's distance at that time = 30 + (1/2)(10) = 35

 

(1/2)x = 35

 

x = 70 mph