# Homework Help: Kinematics 1 Dimension, Person Running Solve for Time

1. Jan 7, 2009

### marindo

1. The problem statement, all variables and given/known data

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 = =

2. Relevant 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

3. 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 dont 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 = =

2. Jan 7, 2009