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Derivation of Distance-Time Gravity Equations
by Ron Kurtus (30 August 2009)
You can derive the gravity equations for the distance an object travels from the starting point with respect to time when the object is dropped, thrown downward or projected upward. You can also find the elapsed time for a given distance.
The equation v = gt + vi, established in Derivation of Velocity-Time Gravity Equations, is used to derive the equation for the distance in a given time. The resulting equation can then lead to the relationship of time for a given distance traveled.
The initial velocity of the object determines the starting direction. If the object is thrown upward, there are three possible times for a given distance from the starting point. One is the time the object takes to go upward some distance, the second is the time it takes to fall back to that distance above the starting point and the third is the time it takes to fall that distance below the starting point.
Advanced Physics and Physical Science students are often required to derive equations. Beginning students who have not yet taken Calculus can skip or skim this lesson.
(See Overview of Derivation of Gravity Equations for a summary of the derivations.)
Questions you may have include:
- What is the distance for a given time equation?
- What is the time for a given distance equation?
- What is the effect of the initial velocity?
This lesson will answer those questions. There is a mini-quiz near the end of the lesson.
Useful tools: Metric-English Conversion | Scientific Calculator.
Distance for a given time
The distance an object travels in a given time is found by knowing that velocity is the change in distance with respect to time:
v = dx/dt
where
- dx is the first derivative of distance x
- dt is the first derivative of time t
Since the equation for the relationship between velocity and time, derived in Derivation of Velocity-Time Gravity Equations, is v = gt + vi, you can substitute that equation for v to get:
dx/dt = gt + vi
Multiply both sides of the equation by dt:
dx = gt*dt + vi*dt
Integrate dx over the interval from 0 to x:
∫ dx = x
where
- ∫ is the integral sign between the two limits
- x is the 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 is:
x = gt2/2 + vit
Positive and negative values
The starting point for dropping or throwing the object is x = 0 and t = 0.
Since the direction of the force of gravity is downward, we consider velocity in that direction as positive and distances below the starting point as positive.
Note: Some text books use a different convention, saying the up is positive. However, that does not seem to follow good logic, since the positive force is toward te ground.
If the initial velocity is upward, vi is a negative number. When the object is above the starting point, x is also a negative number. When the object falls below the starting point, x becomes a positive number.
If the initial velocity is downward, both vi and x are positive.
Obviously, t can only be positive.
Time for a given distance
Derivation of equation for the time it takes for an object to fall a given distance after an initial velocity vi is found by rearranging x = gt2/2 + vit and solving the quadratic equation for t:
x = gt2/2 + vit
Subtract x from both sides of the equation and multiply both sides by 2.
gt2 + 2vit − 2x = 0
Solve the quadratic equation for t:
(See Using the Quadratic Equation Formula in our Algebra section for more information.)
−2vi ± √(4vi2 + 8gx)
t = ________________
2g−2vi ± 2√(vi2 + 2gx)
t = ________________
2g−vi ± √(vi2 + 2gx)
t = ________________
g
Since that is difficult to write on a web page and may not display well on some browser configurations, we will use the following version of the equation, which is also a more compact form:
t = [ −vi ± √(vi2 + 2gx) ]/g
where
- ± means plus or minus
- √ is the square root sign
- √(vi2 + 2gx) is the square root of the quantity (vi2 + 2xg)
- [−vi± √(vi2 + 2gx)]/g is the quantity [−vi± √(vi2 + 2gx)] divided by g
Effect of initial velocity
The direction of the initial velocity influences the time-distance equation.
Thrown downward
When the object in thrown downward, vi and x are positive numbers. Since time can only be a positive number, the following version of the equation is used:
t = [−vi + √(vi2+ 2gx)]/g
Projected upward
When the object is thrown upward, it reaches some maximum height and then falls downward, going past the starting point.
By definition, when vi is going upward, it is a negative number. Thus, −vi is a positive number.
Distances measured above the starting point are negative numbers. Once the object moves below the starting point, x becomes a positive number.
When above starting point
Since the object moves upward and then falls downward, there are two solutions to the time equation for distances above the starting point and the ± sign holds:
t = [−vi ± √(vi2 + 2gx)]/g
For example, x = 0 both before the object is thrown upward and as it passes the starting point on the way down. When x = 0, the equation is:
t = [−vi ± √(vi2)]/g
t = [−vi ± vi]/g
Thus:
t = 0 and t = −2vi/g (remember that −vi is a positive number)
When below starting point
After the object has passed below the starting point, x becomes a positive number. That means that √(vi2 + 2xg) is greater than −vi (a positive number) and the following equation is used:
t = [−vi + √(vi2 + 2gx)]/g
Initial velocity is zero
In the situation where the object is simply falling, vi = 0 and the distance equation is:
x = gt2/2
The time equation is:
t = [√(2gx)]/g
You can simplify the equation by multiplying it by g/√(g2), which equals 1.
t = [√(2gx)]/√(g2)
t = √(2gx/g2)
t = √(2x/g)
Summary
You can derive the gravity equations for the distance an object travels from the starting point with respect to time when the object is dropped, thrown downward or projected upward. You can also find the elapsed time for a given distance.
The equation v = gt + vi is used to derive the equation for the distance in a given time. The resulting equation can then lead to the relationship of time for a given distance traveled.
The initial velocity of the object determines the starting direction.
The derived equations are:
x = gt2/2 + vit
t = [−vi + √(vi2 + 2gx)]/g (for object projected downward
t = [−vi ± √(vi2 + 2gx)]/g (for object projected upward, above the starting point
t = [−vi + √(vi2 + 2gx)]/g (for object projected upward, below the starting point
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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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Derivation of Distance-Time Gravity Equations