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Derivation of Displacement-Time Gravity Equations

by Ron Kurtus (revised 7 January 2011)

The basis for the derivation of the displacement-time gravity equations starts with the equation v = gt + vi that was determined in the Derivation of Velocity-Time Gravity Equations lesson.

Note: Displacement is a vector quantity denoting the change in position in a given direction.

Since velocity is the change in displacement over an increment in time, you use Calculus to integrate that change and get the displacement for a given elapsed time. From that displacement equation, you can then determine the equation for the time it takes for the object to reach a given displacement from the starting point.

The derived equations are affected by the initial velocity of the object. This is important in later applications of the equations.

Questions you may have include:

This lesson will answer those questions.

Useful tool: Metric-English Conversion

Basis for displacement-time derivations

To determine the displacement from the starting point for a given time, start with the equation:

v = gt + vi

(Obtained from Derivation of Velocity-Time Gravity Equations)

where

Note: The initial velocity is the velocity at which the object is released after being accelerated from zero velocity. Initial velocity does not occur instantaneously.

Velocity is also the incremental change in displacement with respect to time:

v = dy/dt

where

By substituting combining these two equations and integrating, you can derive the displacement with respect to time. Then you can rearrange the equation and solve for t to get the time with respect to displacement.

Displacement-time relationship

Displacement-time relationship

Derivation of displacement with respect to time

To obtain the displacement with respect to time, substitute for v in v = gt + vi:

dy/dt = gt + vi

Multiply both sides of the equation by dt:

dy = gt*dt + vi*dt

Integrate dy over the interval from y = 0 to y = y:

∫dy = y − 0

where

Integrate gt*dt over the interval from t = 0 to t = t:

∫gt*dt = gt2/2 − 0

Integrate vi*dt over the interval from t = 0 to t = t:

∫vi*dt = vit − 0

The result of the integrations is the general gravity equation for the displacement with respect to time:

y = gt2/2 + vit

Derivation of time with respect to displacement

You can find the time it takes for an object to travel a given displacement from the starting point by solving the following quadratic equation for t:

y = gt2/2 + vit

Rearrange the equation by subtracting y from both sides of the equation and multiplying both sides by 2.

gt2 + 2vit − 2y = 0

Solve the quadratic equation for t:

t = [ −2vi ± √(4vi2 + 8gy) ]/2g

(See Using the Quadratic Equation Formula in our Algebra section for more information.)

Remove the square root of 4 from inside the square root or radical sign:

t = [ −2vi ± 2√(vi2 + 2gy) ]/2g

The resulting general gravity equation for time with respect to displacement is:

t = [ −vi ± √(vi2 + 2gy) ]/g

where

The plus-or-minus sign means that in some situations, there can be two values for t for a given value of y.

Summary

The basis for the derivation of the displacement-time gravity equations starts with the equation v = gt + vi. Since velocity is the change in displacement over an increment in time, you integrate that change and get the displacement for a given elapsed time.

From that displacement equation, you can then determine the equation for the time it takes for the object to reach a given displacement from the starting point.

The derived equations are:

y = gt2/2 + vit

t = [−vi ± √(vi2 + 2gy)]/g


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Resources and references

Author's Credentials

The following resources provide information on this subject:

Websites

Acceleration due to Gravity Calculations - from Western Washington University

Gravity Resources

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Top-rated books on Simple Gravity Science

Top-rated books on Advanced Gravity Physics


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