Distance over time in a right triangle

[DanaAJames]

I know the Pythagorean theorem for the last part. I am not 100% sure with the other parts. Here is the problem:


Marty and Rediat got in a fight. They walked away from each other on seperate paths at a right angle to each other. Marty walks at 2 km/hr for 20 minutes, then stops to think. Rediat walks 3 km/hr for 30 minutes then he stops to think. Then, suddenly, they both want to apologize and they start running for each other. How far apart are they? How long will it take for them to reach eachother if they run straight at eachother at 5 km/hr?


What I did is I put 20 minutes over 60 minutes to get the fraction of 1/3 for Marty. Then I multiplied that times 2 (because 2 km/hr) and got (1/3 * 2 = 2/3)km.


For Rediat, 30 minutes over 60 minutes, for fraction of 1/2 for Rediat. Then, I multiplied that times 3 (for 3 km/hr) and got (1/2 * 3 = 3/2)km.


Then, since it is a right triangle, I did a²+b²=c²


2/3² + 3/2²= c²


So, 4/9 + 9/4 = c²


√97/36=√c²
√97/36=c

Now, I need to find the square root of 97/36 which is 1.6414763003.

Is this right? If so, how do I complete the problem (including the 5 km/hr part). If not, how do I do it correct?

 

[Ishuda]

Good so far. We have (as you have shown)
(4/9) km² + (9/4) km² = c²
so c = \(\displaystyle \sqrt{\frac{97}{36} km^2} = 1.6414763003\space km \)
and they are c apart. So, if they are c apart, how long will it take them to meet if they are traveling toward each other at 5 km/hr. Note, I would assume that the 5 km/hr is their combined speed and not 5 km/hr each.

 

[Ishuda]

Yes, that's right...good work!

Now, all you need to do is calculate time to travel that @ 10kph

Hah! and Sir Denise has assumed that it is 5 km/hr each. I certainly would not say that was wrong.

 

[HallsofIvy]

I know the Pythagorean theorem for the last part. I am not 100% sure with the other parts. Here is the problem:


Marty and Rediat got in a fight. They walked away from each other on seperate paths at a right angle to each other. Marty walks at 2 km/hr for 20 minutes, then stops to think. Rediat walks 3 km/hr for 30 minutes then he stops to think. Then, suddenly, they both want to apologize and they start running for each other. How far apart are they? How long will it take for them to reach eachother if they run straight at eachother at 5 km/hr?


What I did is I put 20 minutes over 60 minutes to get the fraction of 1/3 for Marty. Then I multiplied that times 2 (because 2 km/hr) and got (1/3 * 2 = 2/3)km.

It might make more sense- to you as well as to us- if you include the units. Marty walks for 20 minutes so for (20 minutes)/(60 minute/hour)= 1/3 hour. He was walking at 2 km/hr so he went (1/3 hour)(2 km/hr)= 2/3 km.

For Rediat, 30 minutes over 60 minutes, for fraction of 1/2 for Rediat. Then, I multiplied that times 3 (for 3 km/hr) and got (1/2 * 3 = 3/2)km.

(30 min)/(60 min/hour)= 1/2 hour. (1/2 hour)(3 km/hr)= 3/2 km.

Then, since it is a right triangle, I did a²+b²=c²


2/3² + 3/2²= c²


So, 4/9 + 9/4 = c²


√97/36=√c²√97/36=c

You might want to write this as √(97/36) to make it clear that the denominator is included in the square root.

Now, I need to find the square root of 97/36 which is 1.6414763003.

Is this right? If so, how do I complete the problem (including the 5 km/hr part). If not, how do I do it correct?

I also interpret "run straight at each other at 5 km/hr" to mean each is running at 5 km/hr. Since they are running at the same speed, they will meet in the middle- each will run 1.64/2= 0.82 km. How long will it take to run 0.82 km as 5 km/hr?

Alternatively, they are "closing" on each other at a relative speed of 5+ 5= 10 km/hr. How long will it take to run 1.64 km at 10 km/hr?

If they were running toward each other at a relative speed of 5 km/h, how long will it take to run 1.64 km at 5 km/hr?

 

[DanaAJames]

I should probably have noted that the problem stated that "How long will it take for the two friends to reach each other if they both start running at 5 km/hr each?"

 

[HallsofIvy]

Yes, the actual statement of the problem, from the start, would have helped!