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Linear Motion (Position, Velocity and Acceleration) Involving Integrals

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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.

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

687 video

video by PatrickJMT

Final Answer

\(t=3.04\sec\)

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A particle has acceleration given by \(a(t)=12t-4\) with initial velocity \(-10\) and initial position \(0\). Find its velocity and position functions.

Problem Statement

A particle has acceleration given by \(a(t)=12t-4\) with initial velocity \(-10\) and initial position \(0\). Find its velocity and position functions.

Final Answer

\(v(t)=6t^2-4t-10\)
\(x(t)=2t^3-2t^2-10t\)

Problem Statement

A particle has acceleration given by \(a(t)=12t-4\) with initial velocity \(-10\) and initial position \(0\). Find its velocity and position functions.

Solution

694 video

video by Krista King Math

Final Answer

\(v(t)=6t^2-4t-10\)
\(x(t)=2t^3-2t^2-10t\)

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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

695 video

video by Krista King Math

Final Answer

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

close solution

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A car braked with a constant deceleration of 16 ft/s2, 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/s2, 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/s2, producing skid marks measuring 200ft before coming to a stop. How fast was the car traveling when the brakes were first applied?

Solution

688 video

video by PatrickJMT

Final Answer

\(v=80 ft/\sec\)

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