Displacement with constant velocity

[clamb2185]

Homework Statement

Two students are traveling along a straight line path. One at .9m/s, the other at 1.9 m/s. How far does the faster student travel in order to arrive at the destination in 5.5 min.

Homework Equations

I was using x=Vx X t for each students path and subtracting the two to determine the distance traveled by the faster student. nope...the solution is 570m. I get 330 m

The Attempt at a Solution

 

[PhanthomJay]

Wecome to the Forums! Is the problem worded correctly? If the faster student is traveling at 1.9 m/s, then in 330 seconds, that student has traveled (1.9)(330) = 627 m, so I don't get either answer. It doesn't matter how fast the other student is traveling, since the problem does not ask for any relative distance between the two students. Are you sure it's 5.5 minutes, not 5.0 minutes?

 

[clamb2185]

Wecome to the Forums! Is the problem worded correctly? If the faster student is traveling at 1.9 m/s, then in 330 seconds, that student has traveled (1.9)(330) = 627 m, so I don't get either answer. It doesn't matter how fast the other student is traveling, since the problem does not ask for any relative distance between the two students.

Hello,

I tried a number of different strategies with this problem. A little background...I am a teacher education department chair and this problem was posed to me by a first year teacher. She is in a small rural school where she is required to teach all science classes in a high school. Physics is not her major.

so...I have lost some sleep on this one over the past 24 hours. At first I thought it was fairly straightforward. I came up with the same answers as you did.

The questions states that two people are walking along a straight path at constant velocity. one at .9 m/s the other at 1.9 m/s. How far would the students have to walk in order for the faster student to arrive 5.5 minutes (330s) before the slower student.

Time zero is when they both depart, one at .9, the other at 1.9. My first year teacher tells me that the answer she is given for this problem is 570 meters??!

I have not been able to reach that solution yet in the various attempts that I have made. I tried using delta x = 1/2 (vfinal) times delta t for each given velocity and that did not hit the mark either...

 

[PhanthomJay]

Sinc this apparently not a homework question, note that the students are both moving at a constant velocity, one at 0.9 m/s, the other at 1.9 m/s. It takes the faster student a Time T1 to reach a certain distance. It takes the slower student a time T2 to reach that same distance. So if the faster student arrives 330 seconds before the slower student, then the slower student's time, T2, is 330 + T1 seconds.

So now you can set up 2 equations

D = 1.9(T1) , where D the distance traveled by the faster student, and
D = 0.9(T1 +330) , where D is the same D as in the first equation.

So set the right side of each equation equal to each other, solve for T1, then solve for D. I get 564 meters. (rounded to 2 significant figures, that's 560 m)

 

[clamb2185]

Thank you Jay, I will send your solution to the teacher. Much appreciated.

Dr. Carmelita Lamb
Turtle Mountain Community College
Department of Teacher Education
Belcourt, ND