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Derivation of Distance-Time Gravity Equations
by Ron Kurtus (revised 28 August 2010)
The basis for the derivation of the distance-time gravity equations starts with the equation v = gt + vi that was determined in the Derivation of Velocity-Time Gravity Equations lesson.
Since velocity is the change in distance over an increment in time, you use Calculus to integrate that change and get the distance for a given elapsed time. From that distance equation, you can then determine the equation for the time it takes for the object to reach a given distance 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:
- What is the basis for the derivations?
- What is the distance for a given time equation?
- What is the time for a given distance equation?
This lesson will answer those questions. There is a mini-quiz near the end of the lesson.
Useful tools: Metric-English Conversion | Scientific Calculator.
Basis for distance-time derivations
To determine the distance from the starting point for a given time, start with the equation:
v = gt + vi
(Obtained from Derivation of Velocity-Time Gravity Equations)
where
- v is the vertical velocity in m/s or ft/s
- g is the acceleration due to gravity (9.8 m/s2 or 32 ft/s2)
- t is the time in seconds (s)
- vi is the initial vertical velocity in m/s or ft/s
Velocity is also the incremental change in distance with respect to time:
v = dy/dt
where
- dy is the first derivative of vertical distance y
- dt is the first derivative of time t
By substituting combining these two equations and integrating, you can derive the distance with respect to time. Then you can rearrange the equation and solve for t to get the time with respect to distance.
Distance-time relationship
Derivation of distance with respect to time
To obtain the distance 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 0 to y:
∫ dy = y
where
- ∫ is the integral sign between the two limits
- y is the vertical distance from the starting point
Integrate gt*dt over the interval from 0 to t:
∫gt*dt = gt2/2
Integrate vi*dt over the interval from 0 to t:
∫vi*dt = vit
The result of the integrations is the general gravity equation for the distance with respect to time:
y = gt2/2 + vit
However, the value of the initial velocity affects the equation.
vi = 0
When the object is simply dropped, the initial velocity is zero (vi = 0) and the equation becomes:
y = gt2/2
vi > 0
When the object is thrown downward, the initial velocity is greater than zero
(vi > 0) and the distance from the starting point is also positive (y > 0).
vi < 0
When the object is projected upward, the initial velocity is less than zero (vi < 0) and the distance y is negative above the starting point—including at the maximum height—and positive below the starting point.
Distance to maximum height
The distance to the maximum or peak height can be derived by starting with the time it takes to reach that height, taken from the Derivation of Velocity-Time Gravity Equations lesson:
tm = − vi/g
or
vi = −gtm
where tm is the time to reach the maximum height.
Substitute for vi in the equation:
y = gt2/2 + vit
ym = gtm2/2 − gtm2
The resulting equation for the maximum height with respect to the time to reach that height is:
ym = −gtm2
Derivation of time with respect to distance
You can find the time it takes for an object to travel a given distance from the starting point by rearranging y = gt2/2 + vit and solving the quadratic equation for t:
y = gt2/2 + vit
Subtract y from both sides of the equation and multiply both sides by 2.
gt2 + 2vit − 2y = 0
Solve the quadratic equation for t:
(See Using the Quadratic Equation Formula in our Algebra section for more information.)
Solve the quadratic equation for t. For convenience, we will use the following version of the equation, which is also a more compact form
t = [ −2vi ± √(4vi2 + 8gy) ]/2g
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 distance is:
t = [ −vi ± √(vi2 + 2gy) ]/g
where
- ± means plus or minus
- √(vi2 + 2gy) is the square root of the quantity (vi2 + 2yg)
The plus-or-minus sign means that in some situations, there can be two values for t for a given value of y.
Again, the value of the initial velocity affects the equation.
vi = 0
When the object is simply dropped, the initial velocity is zero (vi = 0) and the equation becomes:
t = ± √(2gy)/g
Since time t is always positive, the equation is:
t = √(2gy)/g
Change g to √(g2 ) and simplify the equation:
t = √(2gy)/√(g2 )
t = √(2y/g)
t = √(2y/g)
vi > 0
When the object is projected downward, vi and y are positive numbers, and t has only one value. Also, t can only be a positive number. The equation is then:
t = [ −vi + √(vi2 + 2gy) ]/g
vi < 0
When the object is projected upward, the initial velocity is less than zero
(vi < 0). However, that means that −vi is a positive number.
On the way up
Above the starting point, y < 0, and equation used is:
t = [ −vi − √(vi2 + 2gy) ]/g
At maximum height
To find the time it takes to reach the maximum height, rearrange the equation:
ym = −gtm2
tm2 = −ym/g
tm = √(−ym/g)
Since ym < 0, −ym is a positive number.
On the way down
On the way down, y < 0 above the starting point and y > 0 below the starting point. The equation used is:
t = [ −vi + √(vi2 + 2gy) ]/g
Summary
The basis for the derivation of the distance-time gravity equations starts with the equation v = gt + vi. Since velocity is the change in distance over an increment in time, you integrate that change and get the distance for a given elapsed time.
From that distance equation, you can then determine the equation for the time it takes for the object to reach a given distance from the starting point.
The derived equations are:
Distance
y = gt2/2 + vit
ym = −gtm2
Time
t = [−vi ± √(vi2 + 2gy)]/g (general equation)
Object dropped
t = √(2y/g)
Projected downward
t = [−vi + √(vi2 + 2gy)]/g
Projected upward
t = [−vi − √(vi2 + 2gy)]/g (on way up)
tm = √(−ym/g) (at maximum height)
t = [−vi + √(vi2 + 2gy)]/g (on way down)
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Resources
The following resources provide information on this subject:
Websites
Acceleration due to Gravity Calculations - from Western Washington University
Gravity and Gravitation Resources
Books
Top-rated
books on Simple Gravity Science
Top-rated
books on Advanced Gravity Physics
Mini-quiz to check your understanding
If you got all three correct, you are on your way to becoming a Champion in Physics. If you had problems, you had better look over the material again.
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