Kinematics 1 Dimension, Person Running Solve for Time

Click For Summary
SUMMARY

The discussion revolves around a physics problem involving kinematics in one dimension, specifically calculating the time a runner must accelerate to complete a 10,000m run in under 30 minutes. The runner maintains a constant speed of 5.49 m/s for a portion of the race and must accelerate at 0.20 m/s² to cover the remaining 1100m. The key equations used include displacement equations for both acceleration and constant velocity phases, leading to the conclusion that the runner's time of acceleration (T1) and time at constant speed (T2) must sum to 180 seconds.

PREREQUISITES
  • Understanding of kinematic equations, specifically Xf = Xi + Vi(t) + 1/2at²
  • Knowledge of constant acceleration and constant velocity concepts
  • Ability to solve for time and displacement in physics problems
  • Familiarity with unit conversions and basic algebraic manipulation
NEXT STEPS
  • Review kinematic equations for one-dimensional motion
  • Practice problems involving acceleration and constant velocity scenarios
  • Learn how to isolate variables in physics equations effectively
  • Explore additional resources on solving complex kinematics problems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone seeking to improve their understanding of kinematics, particularly in solving time and displacement problems in one-dimensional motion.

marindo
Messages
6
Reaction score
0

Homework Statement



A runner hopes to compleate the 10,000m run in less than 30.0min. After running at a constant speed for exactly 27.0min there are still 1100m to go. The runner must then accelerate at 0.20 m/s^2 for how many seconds in order to achieve the desired time.
*note* person does not accelerate for 1100 m, the person accelerates for a few seconds, then continues running at a constant speed...stupid book = =

Homework Equations



Xf = Xi + Vi(t) + 1/2at^2
Xf = Vi(t)

Vi = 5.49m/s (after calculation)
Vf = 1100m / (180s - T1)

T1 + T2 = 180 seconds
T1 = Time of Acceleration
T2 = Time interval of running at constant speed

S1 + S2 = 1100m

S1 = Displacement during acceleration
S2 = Displacement during constant velocity

The Attempt at a Solution



Vi = 10000m - 1100m (remaining) = 8900m / 1620seconds (3 minutes)
= 5.49 m/s

Vf = 1100 m / [180s - T1]

i basically subbed in the stuff appropriately for S1 and S2 so...

S1 [ (5.5m/s)(T1) + 1/2(0.2m/s^2)(T1)^2] + S2 [1100m/(180 seconds - T1)]x[T2]

i basically got 1100 m to cancel out because i isolated T2 to equal (180 - T1)... so it got rid of that and ...1100 was canceled on both sides = = thus screwing me over T___T

please help... this question is eating at me... and i should know this stuff but i have no clue why i don't T____T i blame the poorly worded question

i also tried doing 1100m = Vi(t) + 1/2a(t)^2 then solving for time = = and to no avail... i got 81 seconds approximately and it was wrong = =
 
Physics news on Phys.org

Similar threads

2-D Kinematic Velocity and Acceleration Practice Problem
  • · Replies 2 ·
Replies
2
Views
2K
How far apart are the stations?
  • · Replies 7 ·
Replies
7
Views
2K
Kinematic problem, are my variables right,
  • · Replies 34 ·
2
Replies
34
Views
3K
How do I calculate work in two-dimensions and find the final velocity and time?
  • · Replies 8 ·
Replies
8
Views
2K
Solving Kinematic Questions: Velocity at t=0
  • · Replies 2 ·
Replies
2
Views
2K
Time it takes for the rocket to fall down
  • · Replies 1 ·
Replies
1
Views
1K
Kinematics Equation Help: Calculating Runner's Acceleration for 10,000m Run
  • · Replies 2 ·
Replies
2
Views
3K
Kinematics - identifying and describing types of motion
  • · Replies 10 ·
Replies
10
Views
2K
Kinematics: Motion in One Direction: Car Chase
  • · Replies 9 ·
Replies
9
Views
3K
Question about Projectile Motion: Throwing a Flower Pot from a Balcony
  • · Replies 18 ·
Replies
18
Views
2K