ACT Problem: Distance, Rate and Time

  • Context:
  • Thread starter Thread starter 816318
  • Start date Start date
  • Tags Tags
    Act Rate Time
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 3K views
816318
Messages
14
Reaction score
0
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?
 
Mathematics news on Phys.org
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".