UNIT 1- KINEMATICS

Kinematics tries to describe how things move – Typically, graphs are used: distance/displacement d versus time t, or

speed/velocity v versus time t graphs can then be analyzed for slopes and areas to give us equations of

motion. Eg. the slope of the distance time graph gives us the formula for speed v = ∆d/∆t.

Unit 1A- Kinematics in One Dimension

The simplest type of motion is constant velocity (uniform motion), an object going at the same speed and same direction. Because direction is important in studying motion, for example whether an object is moving in a straight line or in two dimensions, up or down, toward or away, the concepts of scalar versus vector quantities arises.

Scalar quantities are quantities with only a magnitude and no direction. Eg. Distance d and speed v are scalars eg., 3.0 metres. Other examples of scalar quantities are energy E, work W, and power P.

Vector quantities can have a direction as well as a magnitude. Eg. displacement and velocity . The quantity’s direction is indicated at the end, eg., 3.0 metres West. The direction can also be distinguished by signs +/- for linear motion.The vector versions have a half-arrow over their symbol to distinguish them from their scalar counterparts. Other examples of vector quantities are force F, momentum p and impulse ∆p. The inclusion of direction in vectors can significantly change the answer to a motion problem

Distance d (scalar): the measured length of motion-distance covered, irrespective of direction

Speed v (scalar): the distance covered in a given time t

Displacement (vector): the change in position of an object. (position is where an object is relative to a reference point)

Velocity (vector): the displacement of an object over a given time t.

Example 1: A person walks 5.0 metres forward and then 3.0 metres backwards in 4.0 s.

a) What is the distance the person walked in 4.0 s?

b) What is the displacement of the person’s walk in 4.0s?(use direction and signs)

c) What is the average speed of the person during his walk?

d) What is the average velocity of the person during his walk?

Lab 1 – Graphing : Uniform and Accelerated Motion:

-each 6 dots represents 0.100 seconds. Using a ruler, we can measure the distance between the 6 dots.

DISTANCE vs. TIME Graphs:

Example 2: Perform the lab above as instructed by your teacher. Fill in the table and plot the corresponding total distance d versus total time t graph on graph paper.

Number of Dots Total Time (s) Total Distance (cm)0 0.00 0.00612182430

Example 3: From your distance versus time graph of the cart moving at constantvelocity, calculate the speed v by choosing two points ON

YOUR GRAPH (note: do NOT use any of your original data points as oneof your points) and calculating the slope of your d vs t graph (see next section).

Uniform Motion:

**Speed or Velocity v is the Slope of the Distance versus Time (d vs t) Graph

At any time t, the slope of the distance versus time graph is the speed or velocity. For an object going at constant velocity, the slope of the graph is a straight line. The formula for the slope of a straight line gives us the formula for constant velocity.

v = slope of a distance-time (d vs t) graph

Equation for constant velocity

Because this equation is derived from the formula for the slope of a straight line, as we shall see shortly this formula for velocity v will only hold true for constant velocity.

Examples of Constant Velocity Problems:

Eg 1. Find the velocity (in m/s) for an airplane that leaves Calgary at 9:05am and arrives in Grande Prairie at 11:35am if the two cities are 350km apart.

Eg. 2. How long will it take a bicycle traveling at 8.50m/s to travel 2.35km?

***Eg. 3. Two cars leave Drumheller at the same time for Skullard. One car travels at 80.0km/h and the other at 110km/h. How long will it be before they are 8.00 x 102 km apart?

Eg. 4. An object travels for 5.0s at 15m/s and then changes speed and travels for 100m at 8.0m/s. What was his average velocity?

Should the velocity be changing, then the equation changes to:

Average velocity involves the TOTAL displacement and the TOTAL time. From a graph it is found by the slope of the line joining the final position to the initial position.*vave= slope of a line joining the first to last point on a d vs t graph.

Or =the total displacement over the total time vave=∆d/∆t

What is going to always be true is that for any time t, the slope of the distance-time graph will give us the velocity, irrespective of the shape of the graph (straight or curved).

v = slope of a distance-time (d vs t) graph (always true)

Non-Uniform Motion – Acceleration

Looking at the graph of d vs. t, what is happening to the velocity of the cart over time here? We can’t just take the slope to find velocity! Velocity is changing!

Instantaneous Velocity

The instantaneous velocity corresponds to the slope at any instant of time t. Without using calculus, if we can estimate the slope of the line at any given time t, we can estimate the instantaneous velocity. This is done by drawing a tangent to the curve and finding the slope of the tangent line.

A tangent is a straight line that intersects a curved line at exactly one point – the time point t – and represents the slope of the curved line at that point.

the slope of this tangent line will represent the instantaneous velocity for that point in time t.

Example 10: From your curved distance versus time graph of the cart undergoing accelerated motion, choose a time point t and draw a tangent to the curved line at that point t. Calculate the slope of this tangent line. What does this slope represent?

Constructing Velocity versus Time (v versus t) Graphs

A second motion graph is a plot of the velocity or speed v versus the time t.

For example, the second interval (Interval) runs from dot 6 to dot 12, or a time interval from 0.10 s to 0.20s total time.

The time interval tint = 0.20 – 0.10 s = 0.10 s.

The measured distance for this interval is dint = 36.1 cm – 17.6 cm = 18.5 cm.

(at 0.150 s, midpoint of interval)

From the above, 5 interval speeds can be calculated. The time point chosen for each interval should be the midpoint of the interval; for example, for the second interval 0.10 – 0.20s, the time point chosen should be 0.15 s.

Example 4: Fill in the following table and plot the corresponding interval velocity v versus total time t graph in the accompanying graph paper. For your sample calculations, show the calculation for the examples indicated with an asterisk.

Total Time (s)

Total Distance (cm)

Interval Interval Time(s)

Interval Distance (cm)

Interval Velocity (m/s)

Mid Interval Time (s)

0 00.10 17.6 1 0.0500.20 36.1 2 0.150

0.30 54.9 3 * * * 0.2500.40 72.8 4 0.3500.50 90.4 5 0.450

*sample calculations:tint =

dint =

vint =

Plot a graph of interval velocity vs mid-interval time.

Displacement (Distance) is the Area Under the Velocity (Speed) versus Time Graph

The area under the velocity-time graph represents the distance/displacement of the object. This is always true, irrespective of the shape of the graph (straight or curved).

displacement = area under velocity versus time graph (always true)

For a straight horizontal line velocity versus time graph like the one obtained for uniform motion, the area under the graph is a rectangle. From the formula of the area of a rectangle (area = length x width):

This gives us a second version for the general formula for the displacement d of an object at constant velocity v over some time interval t:

d = v (t) for constant velocity or uniform motion only

Example 5: A cart is traveling a constant velocity of 1.5 m/s. What is the distance it travels between a time of 1.0 s to 3.0 s?

The Area Under an Accelerating Velocity versus Time Graph is a Trapezoid

Remembering that the area under the velocity versus time graph represents the displacement, irrespective of the shape of the graph, we can determine the displacement d over any time interval t by simply calculating the appropriate area, whatever the shape.

Example 11: From the following velocity versus time graph of its motion, calculate the displacement of the moving object between 0 and 3.0 seconds.

The area under the graph here is a trapezoid, which we can break up into a triangle and a rectangle.

d = area under velocity versus time graph

= area of a triangle + area of a rectangle

= ½ (base)(height) + (length)(width)

= ½ (3.0 s)(3.0 – 1.0 m/s) + (3.0s)(1.0s)

= 6.0 m

OR you can use the midpoint of the velocity and multiply by the ∆t: this derives another formula: d = (vf + vi)(t)

2

Example 12: From your own velocity versus time graph of the accelerating cart, calculate the displacement in the first portion of its trip down the ramp.

Example 6: From your graph of interval velocity over time, what is the distance your cart covers during the constant velocity portion of the graph?

Acceleration is the Slope of the Velocity versus Time (v vs t) Graphs

The equation for acceleration (a) also comes from a graph, in this case the velocity (or speed) versus time graph.

a = slope of the velocity versus time (v versus t) graph (always true)

Acceleration is defined, then, as the rate of change of velocity. For an object undergoing uniform motion the slope of the v/t graph is zero (horizontal) and therefore acceleration is zero – what we expect with an object at constant velocity.

a = slope of the velocity-time (v versus t) graph

More commonly, if we call the second velocity v2 the final velocity vf , and the first velocity v1 the initial velocity vi, we derive the general formula for acceleration:

Formula for acceleration:

The units for acceleration are meters per second squared (m/s2).

Example 7: A cart starts from rest at the top of the incline, to a final speed of 1.5 m/sat the bottom of the ramp. The time it took corresponded to 24 dots on a ticker timer, set at a frequency of 60 dots per second (60 Hz). What was the acceleration of the cart on the ramp from the top to the bottom?

Accelerated Motion-Using your v vs t graph:

Instantaneous velocity can be read from a v/t graph instead of using tangents of the d/t graph.

Is there an acceleration of the cart across the counter or floor?? Why?

Equations of Motion are Derived from Motion Graphs

Uniform Motion – Constant Velocity Graphs

(constant velocity formula)

Non-Uniform Motion – Constant Acceleration Graphs

(constant acceleration formula)

From graphical analysis like above, we can derive four more formulas for acceleration problems. The five acceleration formulas are:

a = vf – vi (slope of v/t graph) d = (vf + vi)(t) (area of v/t graph)

t 2

d = (vi) t + ½ (a) t 2 d = (vf) t - ½ (a) t 2

vf2 = vi

2 + 2ad

Using Equations of Motion

When solving motion problems, first decide: is the object moving at constant velocity (uniform motion) or at constant acceleration (non-uniform motion)?

If an object is moving at constant velocity, then only one formula is needed: v = d/t. Values for two of the three variables – the constant velocity v, displacement d, or time t – will be given or known, and the third unknown value can be solved for.

Example 13: The speed of light is a constant in space or air at 3.00 x 108 m/s. How long does it take light to travel around the earth’s 40,000 km circumference?

In constant acceleration problems there typically five possible variables:

acceleration a, initial speed vi, final speed vf, a time interval t, a distance d (and their vector counterparts). All the five acceleration formulas have only four variables: usually you will be given three variables that are known or given values, and the fourth unknown value must be solved for. The correct equation chosen depends on which four (of the five possible) variables are involved in that problem.

Example: A skier starts from rest at the top of a hill and skis to the bottom while constantly accelerating for 10.0 s. If his final speed at the bottom was 20 m/s, what was his acceleration?

This is a (constant velocity/acceleration) problem.

The three given or known variables are: __________________________

The unknown value being looked for is: _______________

The correct equation to use is: _______________

Solve for the unknown value.

Example 14: A sky diver wants to jump of a steep cliff, and knows she needs at least 6.00 seconds to accelerate from rest to a maximum speed of 100 km/h before opening her parachute so it completely opens up and she slowly descends to the ground. Assuming constant acceleration, what is the minimum height of the cliff she is going to need to reach the maximum speed before her chute opens completely?

A constant acceleration problem (not constant velocity)vi =vf = t = d = ?

Example 15: A parked car’s brakes slip and the car slides down a hill, taking 17.0 s before reaching the bottom of the 50.0 m long hill and crashing into a tree. What was the car’s acceleration during its run down the hill?

Example 16: A 200 gram mass is dropped from rest off a tabletop, hitting the floor 1.54 metres below. Ticker tape analysis of the final interval speed just before it hit the floor yielded a value of 5.49 metres per second. What was the acceleration due to gravity for this mass?

Lab 3 - The Free-Fall Problem

Background:

An object that is released into the air, whether thrown or simply dropped, is said to be freely falling once it is released. What happens to objects in free-fall was subject to a lot of debate throughout history.

Aristotle (around 350 BC) believed objects in free-fall fell at a constant velocity. That is, each object picked up speed or accelerated up to a certain maximum speed, then stayed at that speed until it hit the ground. Moreover, different objects had different maximum speeds. A hammer, for instance, fell at a different speed than, say, a feather because of their different weights. Parachutists can attest to this maximum speed reached, which is called terminal velocity.

Galileo (around 1500 AD) believed objects in free-fall fell not at constant velocity but at constant acceleration. Galileo studied the stars out in space. He believed the maximum speeds or terminal velocities reached on earth were due to air resistance, and would not be a factor in space where there is no air. Without air, objects would continue picking up speed (accelerating) indefinitely until they hit the ground. And he believed air resistance also accounted for the differences in the fall rates of differing objects. In other words, Galileo believed a two-pound hammer would fall at the same rate as a feather, except that the feather being very light was slowed down significantly by air resistance. To test this scientifically, he dropped two objects of very different masses (but much heavier than a feather) out of a tower and observed them as they fell to the ground. What do you think he observed?

Materials:Ticker timer and tape2 -200 g masses Cushion pad for mass to land on

Procedure :1) At the edge of your table, using a stand and clamp, attach the ticker timer

vertically as high above the ground as possible. 2) Attach the ticker tape (threaded through timer) by masking tape to the mass

hook and thread it through the timer.3) Switch on timer and drop mass into cushioned box below 4) Repeat with a second ticker tape attaching both masses together and then to

the tape.

Tabulate data to plot a distance-time and velocity-time graph for each mass Plot a distance-time and velocity-time graph (both masses on one graph) From the velocity-time graph slope, calculate the acceleration due to gravity. Who was correct: Aristotle or Galileo? Give reasons from your results.

The Acceleration Due to Gravity ghttps://www.youtube.com/watch?v=E43-CfukEgs

http://www.youtube.com/watch?v=5C5_dOEyAfk

The acceleration due to gravity is a relatively constant value on the surface of the earth for all objects. This constant acceleration is given the symbol g and depends only on the mass of the earth and the distance from the earth’s centre. It does NOT depend on the mass of the accelerating object.

The value for g varies from 9.79 (m/s2) at the equator, to 9.83 at the poles of the earth. This is because the earth is spinning and so is fatter at the equators, and skinnier at the poles. The surface of the earth is slightly farther away from the centre at the equator, and slightly closer at the poles. The intermediate value we use, and assume constant, is:

g = 9.81 m/s2 towards the center of the earth (down)

All freely falling objects on the surface of the earth, then, irrespective of their mass and starting speed, will accelerate at this rate. What it means is that for an object at rest, after one second of falling freely it will be going at 9.81 m/s, after 2 seconds it will be going 19.62 m/s, after 3 seconds it will be going 27.43 m/s and so on. This will hold true for a one kg object as well as a one million kilogram object.

Example 17: An 1.00 kg sandbag is dropped from a hot air balloon, hovering stationary 100 metres above the earth. How long will it take to hit the ground?

Example 18: How fast is an object dropped from a height of 20 metres going when it hits the ground?

Acceleration in One Dimension

Example 22: A dart is shot upwards by a gun at an initial velocity of 14.0 m/s up.a) what is the maximum height the dart will reach?b) how long will it take to reach the maximum height?c) how fast is the dart going as it comes back to its original

height?d) sketch the distance time graph of the dart’s entire trip above.e) Sketch the velocity time graph of the dart’s entire trip above.

Example 23: A rocket 30 metres in the air is moving upwards at 5.0 m/s when it drops its booster straight down. What is the speed of the booster just before it hits the ground 30 metres below?

Because the acceleration due to gravity is quite fast, and back then there were no watches (Galileo was one of the inventors of the pendulum clock), Galileo actually measured the constant acceleration of a ball rolling down an incline or ramp.

Example 26: What is the acceleration of a ball on a ramp rolled upwards initially at 3.0 m/s, let go to go up an incline a short distance and then comes down to its original point of release in 1.7 seconds?

Example 27: A car is moving 100 km/h up a hill when it runs out of gas. The acceleration on this hill is 3.5 m/s2 down. The top of the hill is another 150 m, after which there is a long winding road down that the car can coast to a gas station 2 kilometres away. Will the car make it to the top of the hill?

Example 28: A ball 3.0 m up a ramp is pushed upward at 1.0 m/s and goes 1.0 m up before coming to a stop and rolling back down the ramp. How long does the ball take to go 1.0 m up the ramp and all the way down to the bottom?

Frames of Reference

So far motion of an object has only been considered when the object is moving against a stationary (non-moving) background. But what if the background is moving? For example, what would be the velocity of a baseball if I threw the baseball at 40 km/h West on board an ocean-liner vessel traveling at 30 km/h East?

The answer of the true baseball velocity depends on the observer’s frame of reference that is being considered. For another person traveling on the same ship, the baseball would be observed as traveling 40 km/h West, just as if the baseball was on a stationary background. This is because the observer and I are going the same velocity as the ship, so the ship’s velocity cancels out.

On the other hand, for an observer on the shore or in the ocean, the velocity of the ship must be added to the velocity of the baseball. In this case, the ball is actually traveling at 10 km/h West ( = 40 km/h East + 30 km/h West).

The different frames of reference of an observer onboard the ship and an observer on the ground or ocean produces different velocities for the moving object. Normally we assume the background is stationary unless told otherwise, and consider only the motion of the object relative to that non-moving background – the ground. Yet, in a real sense, all backgrounds are moving: for instance, the earth is itself rotating on its own axis, as well as revolving around the sun so we are moving along with planet earth; likewise the sun is itself moving, as is the solar system, and so on ad infinitum. The problem is not serious as long as we compare the motion of an object relative to a second “non-moving” frame of reference. (Later, when the background becomes an accelerating frame of reference, the problem does become more serious which we will see in circular motion)

Example 29: Pitchers that can throw baseballs at faster than 95 miles per hour –approximately 155 km/h - are highly sought after in professional baseball, usually paid millions of dollars to pitch. My fastest baseball throw is about 100 km/h, yet I can routinely beat the velocity of the fastest pitcher in the major leagues. Can you explain how, and what is wrong about my way of thinking?

Summary of Kinematics in One Dimension

Kinematics is concerned with how things move. The study of motion was analyzed beginning with the measurement of time t and distance d. A graph of constant velocity or uniform motion was generated relating distance versus time (d-t) graph, yielding a velocity v from the slope of the straight line graph of uniform motion. By this analysis, the constant velocity equation is derived v = d/t. In constant velocity problems, two variables of the three (v,d and t) are usually given values, and the third solved for.

A second case of motion was also studied: non-uniform motion or acceleration. Here, the distance versus time (d-t) graph did not yield a straight line, but a curved line difficult to analyze. Instead, the motion was processed into a velocity versus time (v-t) graph that was a straight line. The slope of the velocity versus time graph yields the constant acceleration, while the area under this line gives the displacement.

Moreover, from the straight line v-t graph were derived five equations for cases of motion involving acceleration. Constant acceleration (non-uniform) cases of motion usually involve five variables in total: displacement d, time t, initial velocity vi, final velocity vf and acceleration a. Acceleration problems usually involve only four variables. A typical acceleration problem gives three variables as known values, and the fourth variable- the unknown- is solved for. A summary of the types of graphs and equations for both types of motion – uniform and non-uniform motion – is given below.

Vector Addition of Motion

One-dimensional: Motion in a straight line is one-dimensional motion. Examples of one-dimensional motion are an object thrown straight up or down, a cart moving east and west, or a ball rolling up and down an incline or ramp.

Because motion (displacement and velocity) can be vector quantities, the study of motion often involves vector addition, even for motion in one dimension. The general rule for vector addition is that only vectors going in the same direction can be summed together numerically, so that, for instance, 1 + 1 = 2.

However, all vector problems can be drawn diagramatically! Drawing Vectors: Geographical (Navigational) Coordinates

Vectors can be drawn as arrows, with the length of the arrow corresponding to the vector’s magnitude, and the direction the arrow pointing its direction as defined by the map NESW grid.

Tail-to-Tip Vector AdditionVectors can be added graphically by placing the second vector’s tail to the tip of

the first vector. If there is a third vector, we can add this vector’s tail to the tip of the second vector, and so on. This is called tail-to-tip vector addition. The total vector – called the resultant – is the vector sum of these vectors. Graphically, the resultant is the vector drawn from the origin to the tip of the last vector added tail-to-tip.

Example 19: What is the sum of the two velocity vectors 3 m/s down and 4 m/s down?

If vectors are not in the same direction, then we cannot simply add them together. Instead, we can try and convert one direction to the other so the two directions are the same and can be added together again.

Example 20: Add 3 m/s East to 4 m/s West.

We can choose one direction to be positive, and the other will be the negative for that dimension. For the example above, there are two possibilities: West as positive, East as negative or East as positive, West as negative. Once chosen, the convention must be kept to the conclusion of the problem. Choosing East as positive, West then becomes negative East.

3 m/s East + - 4 m/s East = - 1 m/s East = +1 m/s WestOr we can draw it – scale diagram.

Example 21: What is the sum of the two velocity vectors 5 m/s up and 3 m/s down?

Kinematics Part B - Two Dimensional Motion

Two dimensional motion differs from one dimensional motion in that the motion is not in a straight line. Motion along a plane (flat surface) is two-dimensional motion.

For motion in two dimensions, such as East-West (dimension 1) and North-South (dimension 2), there is no way to convert the one dimension direction into the second to add them. Instead we have to use vector addition with diagrams.

Example 1: Draw a velocity vector of a car moving 25 m/s 75o W of N

We begin by drawing the NESW grid.

Locate the direction

The last step is to draw in the vector as an arrow. This vector arrow is going to begin at the origin (the intersection of the two lines), and extend outward. We can make the size of the arrow any length we want, but since the magnitude of this vector is 25 m/s, we could make the length of this arrow 2.5 cm. Then our scale is simple: 1 cm represents 10 m/s. The tip of the arrow is placed at the end of the 2.5 cm, pointing outward.

Example 2: Draw a velocity vector of 35 m/s at 10o E of S in the above map grid. Use the scale of 1.0 cm = 10 m/s.

Example 3: Add the two velocity vectors 25 m/s 75o W of N and 35 m/s 10o E of S by tail-to-tip vector addition. Draw the resultant velocity vector

Example 4: Add the following two vectors tail-to-tip and sketch in the resultant vector in the following grid.

Vector 1: 3.0 Newtons 30o E of N,

Vector 2: 2.0 Newtons 60o S of E

Trigonometry and Vector Addition

To mathematically sum vectors not going the same direction, we take advantage of the fact that any 2-dimensional vector can be broken down into two components: a horizontal x component and a vertical y component. Typically, for geographic coordinate vectors, we set the convention that the x-component is East-West and East being positive (West is then negative), while the y-component is North-South with North being positive (and South then negative).

Example 5: Break down the velocity vector 1: 25 m/s 75o W of N into x (East-West) and y (North-South) components

.

We can construct a right triangle with the original vector and its x and y components. One of the right triangle angles is 75o and the length of the hypotenuse is equal to 25 m/s. From trigonometry:

therefore, East

therefore, North

From the above we have x = 24 m/s West , y = 6.5 m/s North

Example 6: Break down velocity vector 2: 35 m/s 10o E of S into its x (East-West) and y (North-South) components.

(answer: x = 6.1 m/s East, y = 34 m/s South)

Vector Components: Adding x and y Components Independently

Once vectors have been broken down into (x and y) components, because all similar components are in the same direction, they can be simply summed together. For example, for two vectors, we can add vector 1’s x-component with vector 2’s x-component, and then sum vector 1’s y-component and vector 2’s y-component separately.

Example 7: Find the total x-component and the total y-component from the adding of vector 1: 25 m/s 75o W of N to vector 2: 35 m/s 10o E of S.

X – component Y-componentVector 1: 25 m/s 75o W of N -25sin75 (-24m/s) +25cos 75(+6.5 m/s)

Vector 2: 35 m/s 10o E of S +35sin10 (+6.1 m/s) -35cos 10 (-34 m/s)

Total: -17.9m/s (west) -27.5m/s (south)

After the x-components have been summed up, and the y-components likewise, we can then add the total x-component vectorially (tail-to-tip) to the total y-component. The resultant total vector is identical to the vectorial (tail-to-tip) addition of the two individual vectors. The difference is that this x-component/y-component forms the sides of a simple right triangle. By trigonometry and Pythagoras’ Theorem, the resultant is readily calculated.

Example 8: Add vectorially (tail-to-tip) and sketch the resultant vector of

1) vector 1: 25 m/s 75o W of N and vector 2: 35 m/s 10o E of S 2) the total x-component 17.9 m/s West and the total y-component of 27.5

m/s South of the above.

Example 9: Using trigonometry, calculate the magnitude and direction of the resultant of the above.

The total x-component/ total y-component vector addition gives us a simple right triangle.

From trigonometry, the resultant vector is 33 m/s 57o S of W. Note that the resultant starts from the origin (intersection of the grid lines) and its direction is given as an angle that touches one of the grid quadrants (in this case West), and proceeds to then touch the resultant vector (57o South).

Example 10: Given the following two force vectors: Vector 1: 3.0 Newtons 30o E of N, Vector 2: 2.0 Newtons 60o S of E

Add the two vectors by:

a) separating each vector into its x and y components (see table)b) adding up the total x-component and total y-components (table)c) sketching the vector addition of the total x and total y componentsd) using trig to calculate the magnitude and direction of the resultant

Rectangular (Polar) Coordinate System (RCS)

Because trigonometry is used so much in vector additions, the math convention of the rectangular or polar coordinate system is often a very useful alternative to the geographical conventions NESW.

The rectangular coordinate system starts with 0o as East and proceeds counterclockwise in circular fashion to 360o (East again). North then becomes 90o, West becomes 180o, and South 270o.

Example 11: Sketch the vector 3.0 N 30o E of N, and convert it to RCS degrees.

The advantage of RCS degrees is that the x-components are automatically given as the cosine function, while the y-components are automatically given as the sine function of that angle. There is no longer any need to determine opposite, adjacent, etc.

Example 12: Calculate the x and y component values of 3.0 N 60o RCS

x-component: 3.0 N cos 60o = 1.5 Ny-component: 3.0 N sin 60o = 2.6 N

Once angles have been converted to RCS, vector addition becomes very simple. The sum of all the x-components is the sum of all the cosines (of the RCS angle), while the sum of all the y-component becomes the sum of all the sines (of the RCS angles).

Example 13: Sum the x / y components of 3.0 N 60o added to 2.0 N 60o S of E

First convert to RCS degrees: 3.0 N 60o and 2.0 N 300o RCS

Note how the RCS convention takes care of the positive and negative values of the x and y components automatically (example: sin 300o = -0.866).

Once the total x (x) and total y (y) components are determined, the two component totals can again be added tail-to-tip to give the total resultant vector.

Example 14: Determine the resultant of the total x and total y components above.

Example 15: Add the following two vectors by RCS convention. Sketch the resultant.vector 1: 25 m/s 70o W of N vector 2: 35 m/s 10o E of S.

(Answer: 31 m/s 236o RCS)

Eg1. A jogger runs 1.5 x 103m south, then turns east and runs 1.8 x 103 m. Find the joggers displacement.

Eg2. A car travels east a distance of 30.0 km then E 30o N(30oNof E) for 20.0 km. find the car’s displacement.

Eg3. A plane flies 400km W then turns and flies 200 km N and finally turns W and flies 100kmW. Find the plane’s displacement.

Navigation Problems

Navigation problems are good examples of vector addition problems. In navigation problems, the velocity of the ship or vessel must be added to the velocity of the medium the vessel is riding in. For example, if a boat is moving in a river, the velocity of the boat must be added vectorially to the velocity of the river current.

Example 16: A boat capable of 10 m/s moves in a river with a current of 4.0 m/s east. What is the resultant (true) velocity of the boat when:

a) the boat is heading east at 10 m/s?

b) the boat is heading west at 10 m/s?

c) the boat is heading north at 10 m/s?

d) the boat is heading south at 10 m/s?

Sometimes the resultant velocity is given, for example, when a desired heading is perpendicular to the wind (or current). The real heading of the plane (or boat) is pointed into the wind (or upstream of the current) so that when the plane’s velocity and the wind’s velocity are added up, it produces the resultant heading perpendicular to the wind.

Example 17 : A plane wants to fly at a velocity of 300 km/h straight south (ground velocity) from Edmonton to Calgary. There is a crosswind of 50.0 km/h from the west (in other words, heading east). What is the direction of the plane’s heading (the direction its nose is pointed) and its real speed (air speed)?

Here, the velocity given – 300 km/h south – refers to the desired velocity of the plane, or the resultant. This means the plane’s “true” speed and heading vplane must be added on to the wind velocity vwind 50.0 km/h E to get the resultant desired vplane of 300 km/h S.

(Answer: 304 km/h 9.46o W of S)

Example: The plane wants to fly (ground velocity)300km/h due south. The wind is at 50km/h 35oE of S. What direction does the plane have to head and with what airspeed?

Not only can vectors be broken up into their components, but the components can be thought to act completely independent of each other. So the x and y components of a

two-dimensional problem can be treated for the most part as independent of each other. A problem that involves only the x component can be solved for separately, irrespective for the most part of what’s going on in the y component, and vice versa.

Example 18: A ferry heads across a channel at 10.0 km/h, heading due west. There is a cross-current heading due south at 2.0 km/h. If the channel is 5.0 km wide from east to west, how long does it take the ferry to cross the channel?

Projectile Motion- the Independence of x and y Componentshttps://www.youtube.com/watch?v=D9wQVIEdKh8

Nowhere is the independence of the x and y components from each other more clear than in projectile motion problems.

Projectiles are objects fired in the air. The parabolic arc is due to the x component and y component being quite different and acting independently from each other, but adding together in the end to produce the final path.

The x-component projectile is the horizontal (across) component. Ignoring air resistance, we can treat the x-component as a projectile that is going straight across at a constant velocity. Here the only formula of concern is vx = dx/t.

The y-component projectile is the vertical (up/down) component. Here, it is as if we have a projectile undergoing constant acceleration, the acceleration due to gravity g. We can treat this y-component projectile going up or down as a separate free fall problem, with its own initial velocity vi, displacement dy, final velocity vf, and acceleration g = 9.81 m/s2 down, distinct and acting completely independent from the horizontal motion.

Adding the constant velocity x-component projectile to the free-falling y-component projectile gives the resultant parabolic arc of the projectile. The only factor in common between the two distinct components x and y is the one time t, since in reality there is only one projectile, and that projectile is going to take one time to reach its target.

Horizontal Projectiles

-initial velocity is all x-component and remains constant for all x. -initial velocity in the y direction is zero.

-time can be carried over from one component to the other to solve its problem.

Example 19: A cannon is fired horizontally from the top of a 10.0 metre hill with a muzzle velocity of 200 m/s. What is the horizontal distance the cannonball travels? (dx = ?)

Example 20: A pitcher throws a baseball 100 km/h horizontally off a 30.0 metre high bridge. How far will the baseball travel horizontally before it hits the water below?

(Answer: 68.7 m)

Find the overall final velocity of the object:

Projectiles at an Angle

http://www.design-simulation.com/IP/simulationlibrary/flashsimulations/AirProjectile.swf

-initial velocity is no longer zero for the y-component. -initial velocity can first be broken down into its x-component vx and y-

component vy. -x-component vx is a constant velocity- y-component vy is now the initial velocity vi for the acceleration (free-

fall) problem.

Example 21: A football player throws a football 20.0 m/s at an angle 45.0o to the horizontal. How far away must the (equally tall) receiver be to catch the football?

Note how the acceleration g was changed to -9.81 m/s2 up to keep the directions consistent.

Projectiles at an angle that do not return to the same height can pose an even more difficult problem. In this case, the time t for the y-component may require the formula d=vit + ½at2 and can not be solved directly unless a quadratic formula is used. To avoid having to use a quadratic, we can solve first for the final y-component velocity vf before the projectile hits the ground, then that vf can be used to solve for the time t with another formula.

**max height vf = 0 but time is ½ total time (if returns to same height).**total travel back to same height has a vertical displacement of 0.

Lab 4: Ballistics - Shooting a (Toy) Gun at a Target

Background:

A toy gun with soft plastic bullets is going to be tested and analyzed by teams using nothing more than a metre stick and some masking tape (to mark heights and distances). The goal at the end is that each team gets to shoot at a target setup by the instructor along the counter.

On the floor will be drawn a line which the gun muzzle is to be placed behind. The gun will shoot in the competition from this line.

The perpendicular distance between the gun line and the target will be exactly 4.50 m.

To minimize the 3-dimensional effect and reduce this to a 2-dimensional problem, a row of targets will be lined up along the counter. Knocking over any one of these targets constitutes a direct hit.

During the competition, each team will be allowed 3 trials. During trial number 1 you will be allowed to make measurements and given some time to work out the mathematics before doing the next two trials.

The team knocking the most targets over wins the competition. In the event of a tie, a sudden-death shootout will ensue to determine the winner.

Write-Up (up to three to a team, one write-up per team – diagrams might be helpful)

Show the data values and calculations used to determine the muzzle velocity of your gun by treating it as a horizontal projectile (5 marks)

Show your calculations for shooting at the target in the competition. (5 marks)

Bonus: 1 mark extra for each target hit. Winning team gets 1 extra mark as well.