Kinematics 1 Dimension, Person Running Solve for Time

[marindo]

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 = =

 

[LowlyPion]

The same as:
https://www.physicsforums.com/showthread.php?t=280803

as you have apparently already found out.

 

[thomate1]

I would approach this problem by first understanding the basic principles of kinematics and how they apply to this situation. From the given information, we know that the runner has already completed 27 minutes (1620 seconds) of the 30-minute (1800 seconds) run, leaving 1100 meters to go. We also know that the runner has been running at a constant speed for the first 27 minutes, but then must accelerate at a rate of 0.20 m/s^2 for a certain amount of time in order to achieve the desired time of less than 30 minutes.

Using the equations of kinematics, we can set up the following equations:

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

Where Xf is the final position, Xi is the initial position, Vi is the initial velocity, t is time, and a is acceleration.

We can plug in the given values to solve for Vi, the initial velocity:

Vi = 8900m / 1620s = 5.49 m/s

Next, we can use the equation Xf = Vi(t) to find the final position after the acceleration period:

1100m = 5.49m/s * t + 1/2 (0.20m/s^2) * t^2

Solving for t, we get t = 81.3 seconds.

This means that the runner must accelerate for 81.3 seconds and then continue running at a constant speed for the remaining 98.7 seconds (180 seconds - 81.3 seconds) in order to achieve the desired time of less than 30 minutes.

In conclusion, the runner must accelerate for 81.3 seconds in order to achieve the desired time of less than 30 minutes for the 10,000m run. This is a good example of how understanding the principles of kinematics and applying them to a real-world situation can help us solve problems and make predictions.