## Linear Motion (Position, Velocity and Acceleration) Involving Integrals

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?

$$t=3.04\sec$$

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

$$t=3.04\sec$$

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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.

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

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

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

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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?

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

max height $$144~ft$$; $$t=6~secs$$

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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?

$$v=80 ft/\sec$$

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

$$v=80 ft/\sec$$

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You CAN Ace Calculus

 derivatives integrals

### Trig Formulas

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 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle 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

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

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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