When using calculus for a useful application, the equations and subsequent derivatives usually mean something or describe something. On this page, we discuss the situation when a function represents the position of an object, in two dimension motion, vertically, horizontally or a combination. Calculus, including derivatives and integrals, are the only requirements to understand this material. For a more advanced discussion, see the differential equations page on projectile motion. For a discussion involving only derivatives, see this page. We assume you know the material on the derivatives linear motion page for the following discussion.
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Integration is required when you are given a velocity equation and you are asked to come up with the position function or when you are given acceleration and asked to come up with the velocity and/or the position function. So, essentially, it looks like this.
derivative  position → velocity → acceleration 
integral  position ← velocity ← acceleration 
Of course, with integration, an unknown constant is introduced. So, you will often be given a value that allows you to determine the unknown constant. This value is often called an initial condition. If the value is a position, it may be called an initial position. If it is a velocity, it may be called an intial velocity. You will find the word initial used a lot since most of the time, the value at time \(t=0\) is given.
Let's work some practice problems to get an idea how this works.
Practice
A penny is thrown downward from a 300 foot tall tower with an initial velocity of 50 ft/sec. When does the penny hit the ground?
Problem Statement 

A penny is thrown downward from a 300 foot tall tower with an initial velocity of 50 ft/sec. When does the penny hit the ground?
Final Answer 

\(t=3.04\sec\)
Problem Statement 

A penny is thrown downward from a 300 foot tall tower with an initial velocity of 50 ft/sec. When does the penny hit the ground?
Solution 

video by PatrickJMT 

Final Answer 

\(t=3.04\sec\)
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A particle has acceleration given by \(a(t)=12t4\) with initial velocity \(10\) and initial position \(0\). Find its velocity and position functions.
Problem Statement 

A particle has acceleration given by \(a(t)=12t4\) with initial velocity \(10\) and initial position \(0\). Find its velocity and position functions.
Final Answer 

\(v(t)=6t^24t10\)
\(x(t)=2t^32t^210t\)
Problem Statement 

A particle has acceleration given by \(a(t)=12t4\) with initial velocity \(10\) and initial position \(0\). Find its velocity and position functions.
Solution 

video by Krista King Math 

Final Answer 

\(v(t)=6t^24t10\)
\(x(t)=2t^32t^210t\)
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A ball is thrown straight upward from the ground with initial velocity 96 ft/s. How high does the ball rise and how long does it remain aloft?
Problem Statement 

A ball is thrown straight upward from the ground with initial velocity 96 ft/s. How high does the ball rise and how long does it remain aloft?
Hint 

If you are having trouble getting started on this problem, try using the equation \(v(t)=gt+v_0\) for velocity where \(g\) is the acceleration due to gravity. The negative sign indicates up as the positive direction.
Problem Statement 

A ball is thrown straight upward from the ground with initial velocity 96 ft/s. How high does the ball rise and how long does it remain aloft?
Final Answer 

max height \(144~ft\); \(t=6~secs\)
Problem Statement 

A ball is thrown straight upward from the ground with initial velocity 96 ft/s. How high does the ball rise and how long does it remain aloft?
Hint 

If you are having trouble getting started on this problem, try using the equation \(v(t)=gt+v_0\) for velocity where \(g\) is the acceleration due to gravity. The negative sign indicates up as the positive direction.
Solution 

video by Krista King Math 

Final Answer 

max height \(144~ft\); \(t=6~secs\)
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A car braked with a constant deceleration of 16 ft/s^{2}, producing skid marks measuring 200ft before coming to a stop. How fast was the car traveling when the brakes were first applied?
Problem Statement 

A car braked with a constant deceleration of 16 ft/s^{2}, producing skid marks measuring 200ft before coming to a stop. How fast was the car traveling when the brakes were first applied?
Final Answer 

\(v=80 ft/\sec\)
Problem Statement 

A car braked with a constant deceleration of 16 ft/s^{2}, producing skid marks measuring 200ft before coming to a stop. How fast was the car traveling when the brakes were first applied?
Solution 

video by PatrickJMT 

Final Answer 

\(v=80 ft/\sec\)
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You CAN Ace Calculus
external links you may find helpful 

The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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