Motion

## Big Picture

Everything is in motion. For example, we are all moving around the sun. This means that when we talk about motion, we must look at it relative to something else. The motion of objects through space is one of the first subjects of study for early physicists, but it took a very long time before motion was fully understood. To describe motion, we use rates such a velocity, speed, and acceleration. Throughout this study guide, assume negligible air resistance.

## Key Terms

Displacement: The distance an object has moved from its starting position. SI units: m.

Speed: The rate at which an object covers distance (distance/time). SI units: m/s

Velocity: Speed in a given direction. SI units: m/s

Acceleration: The rate at which an object’s velocity is changing (velocity/time). Like velocity, acceleration is a vector. SI units: m/s^2.

Projectile Motion: The non-linear path an object takes when moving under the constant acceleration of gravity.

Free Fall: Objects falling due to gravity only. Example: an apple falling from a tree is in free fall.

## Describing Motion

### Displacement vs Distance

Displacement  and  distance  both  measure  a  change in  position.  Displacement  is  a  vector  quantity,  while distance is a scalar.

If an object moves in one direction and then returns to its original position, the total distance will be a positive number  and  the  total  displacement  will  be  0.  The diagram below illustrate the concept of displacement. ### Speed vs Velocity

Speed  is  a scalar  quantity  that measures  how  fast an object  is  moving.  In  comparison,  velocity  is  a  vector quantity.  Velocity  is  speed  with  a  direction.  Objects moving at the same speed can have different velocities if they are moving in different directions.

For example, we can say a car is moving at the speed of 25 mph, but if we say the car is moving at 25 mph to the north, then we are stating its velocity.

### Acceleration

Usually  we  use  “acceleration”  to  mean  speeding  up and  “deceleration”  to  mean  slowing  down.  In  physics, acceleration  is  a  vector  quantity  that  could  refer  to speeding up, slowing down, or even changing direction. Careful: positive acceleration does not always mean  speeding  up  and  negative  acceleration does not always means slowing down.

Think  of  an  acceleration  as  a  push.  We  can  define positive acceleration as a push to the right and negative acceleration as a push to the left.

If  an  object  has  a  negative  velocity,  meaning  that  the object  is  moving  towards  the  left,  then  a  positive acceleration pushing the object to the right will cause the object to slow down.

• If  the  velocity  and  acceleration  are  in  the  same direction,  the  object  speeds  up.  If  not,  the  object slows down.

### Instantaneous vs Average

There are two types of speed, velocity, and acceleration: instantaneous and average.

Instantaneous  refers  to a  single  moment in  time,  while average refers to over a period of time.

Motion cont.

## Graphs of Motion

### Distance vs. Time

Below is a graph of distance versus time. The slope $$(\frac{\Delta y}{\Delta x})$$ of the curve represents the speed. As we can see, until $$t = 6$$ s, the object is moving at a constant speed of $$\frac{5m}{3s}$$. After $$t = 6$$s, its speed is 0, meaning the object is at rest. Then, at 11 s, the object starts to move in the opposite direction. After 17 s, it has returned to its original position, resulting in a total displacement of 0. ### Velocity vs. Time

Below is a graph of velocity versus  time.  The  slope represents  acceleration.  If  the  velocity  curve  is  a line  (has  a  constant  slope),  we  know  that  there  is a constant acceleration. The area under the velocity curve is equal to the change in displacement. ### Acceleration vs. Time

Similarly, for acceleration, the area under the acceleration curve is equal to the change in velocity. Below is a graph of acceleration versus time. ## Projectile Motion

Projectile motion is a type of two-dimensional motion where an object moves horizontally (in the x-axis) and vertically (in the y-axis). The x- and y-components of any object’s velocity vector are completely independent of each other. The diagram on the right shows the path and velocity vectors of an object in projectile motion.

• At the top of its fight, when the object stops  rising  and  is  about  to  fall,  the vertical speed of an object is zero.
• Air  resistance  is  ignored,  so  the horizontal  component  of  velocity  stays constant.
• The vertical component of velocity does not  stay  constant  due  to  gravity  -  the object  is  in  free  fall along  the  vertical direction. ## Physics

Motion Problem Guide

## General Guidelines

• Always define a coordinate system for the problem! Define the positive and negative directions.
• Make sure your units make sense.

## Important Equations

• All equations are only valid for constant acceleration
• In  one  dimension,  vectors  can  only  point  in  two directions,  typically  labeled  +  and  -.  The  one dimensional vectors below can be treated like scalars with the signs indicating direction.
Symbol
Meaning
+
-
Substract
X or .
Multiply
$$\div$$ or /
Divide
$$\sqrt{}$$ or $$\sqrt[n]{}$$
square root, nth root
| |
Absolute value
=
Equals
Not equal
Approximately equal
<, ≤
Less than, less than or equal to
>, ≥
Greater than, greater than or equal to
{  }
Set symbol
$$\in$$
An element of a set
( ), [ ]
Group symbols

$$v_{avg} = \frac{\Delta x}{\Delta t}$$

x - displacement

$$a_{avg} = \frac{\Delta v}{\Delta t}$$

v - velociy

$$v = v_0 + at$$

$$v_o$$ - initial velocity

$$\Delta x = v_0t + \frac{1}{2}at^2$$

a - acceleration

$$v^2 = v_0^2 + 2a\Delta x$$

t - time

$$\Delta$$ - change in (final value - initial value)

## Sample Problem

Two  cities  are  33  miles  apart  by  freeway,  where  the speed limit is 65 mph. If it took you 45 minutes to get from one city to the other, what was your average speed?

### Solution

Since we know the distance and the time, we can easily find the speed, which is just distance over time.

$$\text{speed} = \frac{\text{total distance}}{t}$$
$$\text{speed} = \frac{33 mi}{.75 hr}$$
$$\text{speed} = 44 mph$$

## Free Fall

An object thrown into the air is in free fall when it is moving up and falling down. At the maximum height, the object’s velocity is 0 as it changes direction. If the object is at the same height before and after it is thrown, the time it takes to reach the maximum height is exactly half the total time the object spends in the air. The path up and the path down is identical but in opposite directions. Acceleration is always 9.8 m/s2.

Often times, when we solve free fall problems, we look at only one direction (either the journey up or the journey down) because we know the velocity at the top is 0, meaning there is one less variable to deal with.

Equations to use:

$$v = v_0 + at$$

$$v^2 = v_0^2 + 2a \Delta y$$

### Example

A tennis ball is hit up into the air and spends a total of 8 seconds in the air. How high does it reach before coming back down?

### Solution

• Determine what is given and what needs to be found.
• Set up the equation $$v = v_0 + at$$ and solve for$$v_0$$.
• Set up the equation $$v^2 = v_0^2 + 2a \Delta y$$.
• Plug in values and solve for y!

Given: t = 8 seconds

Find: y = ??

Let $$+y$$ be upward and $$-y$$ be the downward direction. In this example, we’re going to use the ball on its way up, so $$t = 4$$ seconds, $$v = 0 m/s$$, and $$a = -9.8 m/s^2$$.

$$v = v_0 + at$$
$$v_0 = v - at$$
$$v_0 = ??$$
$$v_0 = .8 m/s^2 . 4 s$$
$$v_0 = 39.2 m/s$$

$$v^2 = v^2 + 2ay$$
$$y = \frac{v^2 - v_0^2}{2a}$$
$$y = ??$$
$$y = \frac{-(39.2 m/s)^2}{-2.9.8 m/s^2}$$
$$y = 78.4m$$

So the ball reaches a height of 78 m.

## Physics

Motion Problem Guide cont.

## Projectile Motion

In projectile motion, the acceleration in the y-direction is always -9.8 m/s2, while the acceleration in the x-direction is 0. If the object is thrown at an angle, the initial velocity must be broken up into x- and y-components ($$v_x = v \cos \theta, v_y = v \sin \theta$$). Like in freefall, the velocity in the y-direction at the top of the object’s path is 0 m/s.

In some  questions, the object is thrown  at a specific angle “above the horizontal”  or “below the horizontal.” In the example and diagram below, you can see what “above the horizontal” means. “Below the horizontal” is similar, only vy and acceleration are both in the same direction.

Equations to use:

$$v = v_0 + at$$

$$\Delta x = v_0t + \frac{1}{2}at^2$$

### Example

A man standing on a cliff throws a rock at 25° above the horizontal with an initial velocity of 15 m/s. If the rock is in the air for 5 seconds, how tall is the cliff, and how far into the ocean does it travel?

### Solution

• Determine what is given and what needs to be found.
• Draw and label a diagram.
• Set up the equation $$\Delta x = v_0t + \frac{1}{2}at^2 for the x-and y-directions. • Plug in values and solve! Let the initial positions \(x_o$$ and yo be $$0$$ so that $$Δx = x - xo = x \text{and} Δy = y - yo$$.

Given: $$v_o = 15 m/s$$, $$t = 5$$ seconds, and $$θ = 15°$$
Find: $$x$$ and $$y$$

x - direction

y - direction

$$x = v_{0,x^t} + \frac{1}{2}a_xt^2$$

$$y = v_{0,y^t} + \frac{1}{2}a_yt^2$$

$$v_{0,x} = v_0 \cos \theta$$

$$v_{0,y} = v_0 \sin \theta$$

$$a_x = 0 m/s^2$$

$$a_y = -9.8 m/s^2$$

$$x = 15 m/s . cos(15^\circ)$$

$$y = 15 m/s . sin(15^\circ) - \frac{1}{2} . 9.8 m/s^2 . (5 s)^2$$

$$x = 68.0 m$$

$$y = -90.8 m$$

So the cliff is 91 m tall, and the rock travels 68 m.  In this problem, we end up with a negative y value that represents the height of the cliff because y = 0 is at the top of the cliff.

### Example

If an object is launched into the air with speed  at an angle  on flat ground, how long will the object be in the air?

### Solution

Since we know that the object’s final vertical displacement is zero, we can think of its path as a parabola. We can find the roots using the displacement equation.

$$\Delta y = v_0t + \frac{1}{2}at^2$$

Start with the equation for vertical displacement.

$$0 = v_y sing \theta t - \frac{1}{2}gt^2$$

Substitute in known values. The sign changed on the second term because we know g is negative.

$$0 = t(v_y \sin \theta - \frac{1}{2}gt)$$

Begin solving for t like we would for any quadratic.

$$t = 0, \frac{2_vy \sin \theta}{g}$$

We already know the displacement is 0 at time t = 0, so the second solution is our answer.