MHB ACT Problem: Distance, Rate and Time

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SUMMARY

The discussion focuses on solving a distance, rate, and time problem involving two cyclists, Joan and Anthony, using the formula d=rt. Joan rides at 15 miles per hour while Anthony rides at 12 miles per hour. It is established that Joan will catch Anthony in 5 hours, leading to the conclusion that Anthony is 15 miles ahead of Joan. The problem is solved by setting up equations for their positions over time and equating them at the point of intersection.

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  • Understanding of the distance formula d=rt (distance equals rate times time)
  • Basic algebra for solving equations
  • Familiarity with functions and graphing
  • Knowledge of relative motion concepts
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  • Learn how to set up and solve equations involving multiple variables
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Students preparing for the ACT, educators teaching algebra concepts, and anyone interested in applying mathematical principles to real-world scenarios involving motion.

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Joan is riding her bicycle along a track at 15 miles per hour. Anthony, who is ahead of Joan on the same track, is riding his bicycle at 12 miles per hour. If it will take Joan 5 hours to catch Anthony at their current speeds, how many mile ahead of Joan on the track is Anthony?

How would you solve it using the d=rt formula?
 
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Re: ACT problem

816318 said:
Joan is riding her bicycle along a track at 15 miles per hour. Anthony, who is ahead of Joan on the same track, is riding his bicycle at 12 miles per hour. If it will take Joan 5 hours to catch Anthony at their current speeds, how many mile ahead of Joan on the track is Anthony?

How would you solve it using the d=rt formula?

Hi 816318, could you expand on what you intend the d = rt formula to mean? Maybe i should know from experience.. I'm thinking distance equals something by time.. Ha :p. I'll feel silly when i realize, but we have to know for sure!
 
Re: ACT problem

816318 said:
Joan is riding her bicycle along a track at 15 miles per hour. Anthony, who is ahead of Joan on the same track, is riding his bicycle at 12 miles per hour. If it will take Joan 5 hours to catch Anthony at their current speeds, how many mile ahead of Joan on the track is Anthony?

How would you solve it using the d=rt formula?

We can simplify this problem a bit if we orient our coordinate axis such that Anthony is at the origin and Joan is some distance away approaching the origin at 3 mph. Can you proceed?
 
Re: ACT problem

MarkFL said:
We can simplify this problem a bit if we orient our coordinate axis such that Anthony is at the origin and Joan is some distance away approaching the origin at 3 mph. Can you proceed?

Thanks I got it now, d=3(5) 15!
 
Re: ACT problem

Another approach would be to initially put Joan at the origin and Anthony at $d$. Disnaces are in miles and time in hours. And then:

Joan's position as a function of time is:

$$J(t)=15t$$

Anthony's position as a function of time is:

$$A(t)=12t+d$$

Now, we are told they meet in 5 hours, or:

$$J(5)=A(5)$$

$$15(5)=12(5)+d$$

$$d=15(5)-12(5)=3(5)(5-4)=15$$
 
Re: ACT problem

Joppy said:
Hi 816318, could you expand on what you intend the d = rt formula to mean? Maybe i should know from experience.. I'm thinking distance equals something by time.. Ha :p. I'll feel silly when i realize, but we have to know for sure!
distance traveled= rate of travel times time traveled.

You may know it better as "d= vt" where "v" is now "velocity".
 

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