How Long to Accelerate for Desired Time?

Thread starter pointintime
Start date Aug 20, 2009
Tags

Motion One dimensional One dimensional motion

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SUMMARY

The discussion focuses on calculating the time a runner must accelerate at 0.20 m/s² to complete a 10,000 m run in under 30 minutes. After running for 27 minutes, the runner has 1100 m left and an initial velocity of 5.494 m/s. The participants derive equations based on kinematics, specifically using the formulas for displacement and acceleration, to find the time required for acceleration. The final equations lead to a quadratic form that can be solved for the acceleration time.

PREREQUISITES

Understanding of kinematic equations, specifically displacement and acceleration formulas.
Familiarity with average velocity and its calculation.
Ability to manipulate algebraic equations and solve quadratic equations.
Knowledge of unit conversions, particularly between minutes and seconds.

NEXT STEPS

Study the derivation of kinematic equations, focusing on displacement and acceleration.
Learn how to solve quadratic equations and apply them in real-world scenarios.
Explore the concept of average velocity and its implications in motion problems.
Practice unit conversion techniques, especially in physics-related contexts.

USEFUL FOR

Students studying physics, particularly those focusing on kinematics, athletes analyzing performance metrics, and educators teaching motion concepts.

Aug 21, 2009

pointintime

I don't know how to get this mess into standard form...

-2^-1 t3^2 - 180 s (t3) + t3^2 = a^-1 (Vo (180 s) - 1100 m + Xo)

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Aug 21, 2009

pointintime

I have to get it into

y = ax^2 + bx + c

don't know how

Aug 21, 2009

pointintime

rewrote for zero don't know what to do

-2^-1 t3^2 - 180 s (t3) + t3^2 = a^-1 (Vo (180 s) - 1100 m + Xo)
(Vo (180 s) - 1100 m + Xo)^-1 a(-2^-1 t3^2 - 180 s (t3) + t3^2) = 0

:O