Speed Up to Catch Leader in 10km Race

  • Thread starter Thread starter negation
  • Start date Start date
  • Tags Tags
    Kinematics Motion
Click For Summary

Homework Help Overview

The discussion revolves around a problem in kinematics related to a 10km race where a runner is attempting to catch up to a leader. The original poster presents a scenario where they are 100m behind the leader after 35 minutes of running, seeking to determine the necessary acceleration to reach the leader by the finish line, assuming the leader maintains a constant speed.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations of speeds for both runners and the time required for the slower runner to catch up. There is an emphasis on ensuring clarity in variable usage and the implications of using the same variable for different runners.

Discussion Status

Some participants have provided calculations and expressed confusion regarding the results, particularly in relation to the answer sheet. There is a recognition of the need to apply kinematic equations to find the required acceleration, and a suggestion to explore the SUVAT equations has been made. The discussion reflects a mix of attempts and clarifications without reaching a consensus.

Contextual Notes

Participants note the significance of the distances each runner has left to cover and the time constraints involved. There is an acknowledgment of the original poster's concern about the time spent on the problem and the potential impact on their well-being.

negation
Messages
817
Reaction score
0
After 35mins of runnin, at the 9km point in a 10km race, you find yourself 100m behind the leader and moving at the same speed.
What should your acceleration be if you're to catch up by the finish line? Assume that the leader maintains constant speed.
 
Last edited:
Physics news on Phys.org
Please show your own attempt.
 
vf = 9000m/35min = 257m min^-1
vf = 9100m/35min = 260 min^-1

10,000-9100m = 900m
900m/ 260m min^-1 = 3.46min

Runner requires 3.46min to tie the leader at the finish line.

xf - xi = 0.5(vf + vi)t
1000m = 0.5(vf+257m min^-1)3.46min
vf = 321m min^-1

vf doesn't tally with the answer sheet
 
Can someone help me? I've spent 4 hrs on this question already. I don't want to get into the unhealthy habit of not sleeping for days until I figure out the answer.
 
The numbers that are significant in this problem are the 1000 m that the slower runner needs to cover, and the 900 m that the faster runner has left.

Don't use the same variable for both runners:
negation said:
vf = 9000m/35min = 257m min^-1
vf = 9100m/35min = 260 min^-1
vf can't possibly be equal to two different numbers.
 
vf = 9000m/35min = 257m min^-1
vf = 9100m/35min = 260 min^-1

10,000-9100m = 900m
900m/ 260m min^-1 = 3.46min

I agree. You have to get to the finish in less than 3.46min.

Than I would look at one of the SUVAT equations to work out the minimum constant acceleration required. http://en.wikipedia.org/wiki/Equations_of_motion

Perhaps this one and solve for a...

s = ut + 0.5at2

You know..

s = 900m
u = 257m/min or 4.28m/S
t = 3.46min or 207.6 seconds
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
4K