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Linear Motion (Position, Velocity and Acceleration) 

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. Beginning calculus topics are the only requirements to understand this material. If you've studied integration, even better. However, don't worry if you haven't. You can still understand most everything on this page. Just skip the parts with integration and come back here when you have. For a more advanced discussion, see the differential equations page on projectile motion. 

Usually we derive or start with a function whose variable is time. This usually looks like \( s(t) \) where the independent variable is t and represents time. The function itself is called a position function. This means that we can plug in a time, like \( t=3 \), to the function, \( s(3) \) and the result is the ( instantaneous ) position of the object at time \( t=3 \). Of course the units are dependent upon what the equation represents. Also, we usually assume that time starts at \(t=0\) and always increases.
When we have a position function, the first two derivatives have specific meanings. The first derivative is the velocity and the second derivative is the acceleration of the object. We take the derivative with respect to the independent variable, t.
The units of velocity are distance per unit time, in MKS units, meters per second, m/s.
The units of acceleration are distance per unit time squared, in MKS units, meters per second squared, m/s^{2}.
Note About Velocity   There is a difference between velocity and speed. Velocity can be negative and includes direction information. Speed is the magnitude of velocity and is always positive. Speed does not include direction. If you are familiar with vectors, velocity is a vector, speed is a scalar whose value is the magnitude of a velocity vector. Many people use these terms interchangable, which is incorrect.
Here are the equations you need.
Equations  

position  \(s(t) = at^2/2 + v_0t + s_0\)  distance 
velocity  \(s'(t) = v(t) = at + v_0\)  distance per unit time 
acceleration  \(s''(t) = a(t) = a\)  distance per unit time squared 
Note About Acceleration   In the above table, we assumed acceleration is constant. This is true in many problems, especially ones where we talk about the acceleration due to gravity near the surface of the earth, for example. However, this is not true in ALL problems. So you need to pay attention to what is going on in the problem statement if you are required to set up these equations.
Constants and Variables  

\(s_0\)  initial position; can also be written \(s(0)\) 
\(v_0\)  initial velocity; can also be written \(v(0)\) 
\(a\)  acceleration is often, but not always, constant 
\(t\)  independent variable time 
Okay, let's watch a video to get this into our heads a little bit more. Here is a good video that starts out by giving an overview of the derivative relationship between these equations. Then he does a quick example, and then finishes by giving a more intuitive description of the relationships. He does talk a little bit about antiderivatives but if you aren't there yet, this video will still help you.
PatrickJMT  Position, Velocity, Acceleration using Derivatives  
Your best course of action now is to work as many practice problems as possible. These problems require only differentiation.
Practice 1  

\(s(t)=3t^2+5t+2\) describes a particle's motion with units in meters. Find
a. the velocity and acceleration functions, and  
answer 
solution 
Practice 2 

A particle moves along a path with its position \(p=t^26t+8\) meters. a. When does the velocity equal 30 meters per second? 
solution 
Practice 3  

If the position of a particle is given by \(x(t)=10016t^2\), find the position of the particle when the velocity is zero.  
answer 
solution 
Practice 4  

A car with position function \(x(t)=100t5t^2\) is traveling \(100~ft/s\) when the driver suddenly applies the brakes. How far and for how long does the car skid before it comes to a stop?  
answer 
solution 
Practice 5  

The vertical position of a ball is given by \(y(t)=16t^2+96t+50 ft\). What is the maximum height the ball will reach?  
answer 
solution 
Practice 6  

A ball with position function \(y(t)=gt^2/2+v_0t+s_0\) is thrown straight upward from the ground with an initial velocity of 96 ft/sec. Find the maximum height that the ball attains and its velocity when it hits the ground.  
answer 
solution 
Practice 7  

The motion of a ball is described by \(y(t)=16t^2+64t\). Find the maximum height and the velocity when the ball hits the ground.  
answer 
solution 
Practice 8 

A coin is dropped from the roof of a 600ft tall building with initial velocity of 8 ft/sec. When does it hit the ground and what is its velocity at that point? 
solution 
Using Integration 
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 9  

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?  
answer 
solution 
Practice 10  

A particle has acceleration given by \(a(t)=12t4\) with initial velocity \(10\) and initial position \(0\). Find its velocity and position functions.  
answer 
solution 
Practice 11  

A ball is thrown straight up from the ground with initial velocity \(v_0=96\) ft/sec. Find the ballâ€™s maximum height and its velocity when it hits the ground.  
answer 
solution 
Practice 12  

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?  
answer 
solution 