Kinematics in Two Dimensions; 2D- Vectors

Kinematics in Two Dimensions; 2D- Vectors

Addition of Vectors Graphical Methods Below are two example vector additions of 1-D displacement vectors. For vectors in one dimension, simple addition and subtraction are all that is needed. You do need to be careful about the signs, as the figure indicates. The resultant vector is the vector sum of the two vectors being added

If the motion is in two dimensions, the situation is somewhat more complicated. Here, the actual travel paths are at right angles to one another; we can find the displacement by using the Pythagorean Theorem. The angle can be found by using trig. D tanθ = 2 D 1

Adding the vectors in the opposite order gives the same result:

The relationship on the previous slide (see below), is generally called a vector equation or relationship. The vectors D D R = D 1 + D 1 and D 2 were added using the tip to tail or 2 head to tail method to form a D R vector diagram (on left). The D 2 resultant vector is drawn from the tail of the first vector to the D 1 head of the final vector. In the previous example, to find the magnitude of the vector, the Pythagorean theorem could be used because the vector diagram is a right angled triangle. What happens if the two vectors being added are not at right angles?

If two vectors being added are not at right angles it takes the Cosine and Sine laws to solve for resultant vector. For example, in the situation below. The lengths of the triangle are related to the magnitude of the vectors, and the angles can give direction. The relationship d 3 = d 1 + d 2, has a θ corresponding vector diagram. Cosine and Sine Laws: c 2 = a 2 + b 2 2abcosC A c B b a C sin A a = sin B b d 1 = sinc c d 3 d 2

Here is another example of vectors being added graphically by using the tail-to-tip or Head to tail method. In this case we are adding 3 vectors rather than the usual two. The vector relationship is: V R = V 1 + V 2 + V 3

Other points about vectors The direction of a 2-d vector is usually given by an angle relative to the x or y axis Another way to indicate direction is by using bearings format. One uses N,S,E, and W, and indicated by making reference to these directions. A 2-D vector is generally represented by a positive magnitude (no such thing as a negative magnitude for vectors) and a direction. W N S y E x

Example 1: Solution: 5 km, 37 0 N of E.

Example 2: A hiker walked 9.00 km 30 0 W of N at 3.00 km/h, then walked 16.0 km 20 0 N of W at 2.00 km/h. a) What was his final position (displacement)? b) What was the duration of the trip? c) Calculate his average speed and average velocity over the entire trip.

a) First define your position vectors and state a vector relationship: Solution: d = 9.00 km 30.0 0 W of N = 16.00 km 20.0 0 1 d2 N of W d 3 =? d 3 = d 1 + d 2 Now draw the vector diagram: d 2 Using cosine law: d 32 = 9 2 +16 2-2(9)(16)cos(140 0 ) = 557.62 d 3 = 23.6 km d 3 And using sine law: sinθ 16 = sin(140) θ = 25.82 0 23.61 Finally, d 3 = 23.6 km 55.8 0 W of N 20 0 d 1 30 0 30 0 ϴ

b) t= 9/3 + 16/2 = 2 + 8= 11 h c) Average speed = (total distance travelled)/ time = 25 km/ 11 h= 2.27 km/h Average velocity = (change in displacement)/ time υ = 2.15 km/h 55.8 0 W of N = 23.6 km/ 11 h= 2.15 km/h Notes: The average velocity is a vector, so it has a direction. It shares the same direction as the net change in position (displacement) vector. When adding vectors head to tail it is useful to draw the dotted axis lines (see dotted blue lines on previous slide). This helps to determine the angles in the vector diagram. Normally you should give the final direction of a vector relative to the x or y axis.

Adding Vectors by Components Any vector can be expressed as the sum of two other vectors, which are called its components. Usually the other vectors are chosen so that they are perpendicular to each other (x and y component vectors).

If the components are perpendicular, they can be found using trigonometric functions.

The components are effectively one-dimensional, so they can be added arithmetically:

In summary, if the the vector relationship were: V = V 1 + V 2 Then, V x = V 1x + V 2 x and V y = V 1y + V 2 y The x component of vector is equal to sum of the x components of the other two vectors. A similar statement could be made about the y components. V

Adding Vectors by Components Adding vectors: 1. Draw a diagram; add the vectors graphically. 2. Choose x and y axes. 3. Resolve each vector into x and y components. 4. Calculate each component using sines and cosines. 5. Add the components in each direction. 6. To find the length and direction of the vector, use:

Example 3: In this example we will use the same vectors added in example 1 (slide 10) but add using components. d = 9.00 km 30.0 0 W of N, = 16.0 km 20.0 0 1 d2 N of W. Find d3 where: d 3 = d 1 + d 2 Solution: 1) First find the components of both vectors d 2 d 1x 30 0 d 1y d 2y 20 0 +x d 2x d 1 d 1x = -(9)(sin30)= -4.5 d 1y = (9)(cos30)=7.794 d 2x = -(16)(cos20)= -15.04 d 2y = (16)(sin20)= 5.472 d 3x =d 1x +d 2x = -19.5351 km d 3y =d 1y +d 2y = 13.267 km 3) Add the x and y components of d 3 (head to tail) to find its magnitude and direction: d 3 = ( 19.5351) 2 + (13.267) 2 = 23.6 km 13.27 d 3 19.54 +y +x tan -1 (13.267/19.5351) = 34.2 0 2) Find the components of d 3

Subtraction of Vectors, and Multiplication of a Vector by a Scalar In order to subtract vectors, we define the negative of a vector, which has the same magnitude but points in the opposite direction. Then we add (Tail to Tip method) the negative vector:

Subtraction of Vectors, and Multiplication of a Vector by a Scalar A vector V can be multiplied by a scalar c; the result is a vector cv that has the same direction but a magnitude cv. If c is negative, the resultant vector points in the opposite direction.

A vector subtraction can be written as an addition. For instance perhaps one is interested in finding, where: d 3 d 3 = d 2 d 1 One could rewrite the vector relationship as: d 3 = d 2 + ( d 1 ) Realizing that d 1 is in the opposite direction to. One can just add d 2 and d 1 head to tail to find d3. d1 Yet another option is to add d 1 to both sides of the original vector relationship which gives: d1 + d 3 = d 2 I have done a fully worked example on a vector subtraction looking at the above options and by using components to do the subtraction. Please see Vector Subtraction example.. in the notes section. If you use any of these techniques please follow the method I have demonstrated. A common time to perform a vector subtraction could be in the following: relative velocities, change in velocity, and change in momentum.

Projectile Motion(2-D) A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

It can be understood by analyzing the horizontal and vertical motions separately.

The speed in the x-direction is constant; in the y- direction the object moves with constant acceleration g. This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

If an object is launched at an initial angle of θ 0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component.

Solving Problems Involving Projectile Motion Projectile motion is motion with constant acceleration in two dimensions, where the acceleration is g and is down.

1. Read the problem carefully, and choose the object(s) you are going to analyze. 2. Draw a diagram. 3. Choose an origin and a coordinate system. 4. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g. 5. Examine the x and y motions separately.

6. List known and unknown quantities. Remember that v x never changes, and that v y = 0 at the highest point. 7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

Projectile Motion Is Parabolic In order to demonstrate that projectile motion is parabolic, we need to write y as a function of x. When we do, we find that it has the form: This is indeed the equation for a parabola.

Relative Velocities Velocities like anything else are measured relative to the observer. The motion of an object will (could) have a different velocity relative to an observer in different reference frame. To state a velocity there is a certain useful notation where the object and frame are indicated. A common reference frame is the ground frame (g). This is the reference frame where one is stationary relative to earth when making an observation. For example if one said the velocity of object A was 10.0 m/s north one could write: υ A/g = 10 m/s North The observer (in the ground frame) observed the velocity to be 10 m/s north.

1-D Relative Velocities One can gain an understanding of the mathematical process of determining relative velocities by examining the 1- d case. Example: a) Object A is travelling to the right at 2.00 m/s and object B is travelling to the left at 3.00 m/s. What is object A s velocity relative to object B ( V A/B )? Or in other words if you were an observer on object B, what velocity would you measure object A s to be? Let the positive x direction be to the right. +x Answer: One can see intuitively it would be 5.00 m/s in the positive x direction Answer: On can see it would be -5.00 m/s A v A/g = 2.00 m/s v B/g = 3.00 m/s b) What is object B s velocity relative to object A ( V B/A )? B

In the previous example what vector operation was applied? One can see that there was a vector subtraction. In part a) one was trying find the object A s velocity relative to object B. One in effect wishes to put object B at rest (subtract B s velocity). The vector operation for a) is: VA/B = V A/g V B/g = 3-(-2) = 5 m/s In part b) one is trying to find object B s velocity relative A. Therefore, V B/A = V B/g V A/g = -3-2 = -5 m/s Example: Object A was travelling left at 7.00 m/s and object B was also travelling left at 3.00 m/s. What is the velocity of object B relative to A? Solution: V B/A = V B/g V A/g = -3-(-7) = 4 m/s +x The same mathematical rules apply to relative velocities in two dimensions.

A cool problem which is much more easily answered using relative velocities. Two trains (initially 40 km apart) approach each other on a collision course. One train is travels east at 20 km/h while the other travels west at 45 km/h. Meanwhile a fly left the front of one train and travels back and forth from the fronts of both trains until the the trains collide. If the fly travels with an average speed of 50 km/h, what total distance would the fly travel? Solution: Use relative velocities to determine the elapsed time till the trains collide. t= distance/velocity = 40 km/ 65 km/h = 0.6154 h The distance the fly travelled: vt= (50 km/h)(0.6154 h) = 31 Km

2-D Relative Velocities Example: A Boeing 747 has a velocity of 650 km/h north, while a Cessna 172 has a velocity of 250 km/h east. What is the velocity of the Boeing relative to the Cessna? 1) Define your vectors and state a vector relationship first 2) Construct a vector diagram 3) Solve triangle for desired vector. Solution V A/g = 650 km/h N V B/g = 250 km/h E V B/g = 250 km/h W V A/B =? Construct the vector diagram: v B/g = 250 Km/h V A/B = V A/g V B/g V A/B = [ 650 2 + 250 2 ] 0.5 or = 696 km/h V A/B = V A/g + ( V B/g ) v A/B v A/g = 650 km/h Θ= tan -1 [250/650]= 21.0 0 One could have use vector components to complete the vector subtraction and find the same result.

In the previous example the original vector relationship could be restated as follows V A/B = V A/g V B/g VB/g + V A/B = V Adding V B/g to both A/g sides gives: Vector relationship A Vector relationship B The vector diagram based on this vector relationship (B) is give by: v A/B Solving this vector triangle will give the same result as on the previous slide. v A/g = 650 km/h The subscript notation that has been introduced can be used to verify the vector relationship: v = 250 Km/h B/g Examining the subscripts of relationship A (subtraction means divide subscripts): A B = A g B g Examining the subscripts of relationship B (addition means B A = A g B g multiply subscripts): One can see that the left hand side equals the right hand side in each case, verifying our relationships are set up correctly according to our definitions of relative velocities.

Example: A boat travels at 11.11 m/s in still water. It aims directly across a river (800 m wide). There is a river current of 7.50 m/s to the left. The river will inevitably carry the boat downstream. a) State the vector relationship relating the vectors: the boat s velocity relative to the water v b/w, the water s velocity relative to ground v w/g, and the boat s velocity relative to the shore (ground) v b/g 800 m b) Sketch the vector diagram relating the three vectors. c) Determine the boats velocity relative to the shore. v b/g d) How long does it take the boat to reach the other side of the river? e) At what angle should the boat aim upstream in order to move straight across the river (relative to shore)? What is the velocity of the boat relative to the shore in this case? Redraw the vector diagram. f) How long will the boat take to cross the river in this latter case?

Solution: v b/w a) v w/g v =? b/g 1 st List what you know about the vectors = 11.11 m/s straight across = 7.50 m/s left v b/w + v w/g = b/g 2 nd State the vector relationship: v 800 m v w/g v b/g b) v b/w c) v b/g = (7.50) 2 + (11.11) 2 = 13.4 m/s θ = tan 1 7.50 = 34.0 0 11.11 d) t= (800 m)/( 11.11 m/s) = 72.0 s 3 rd Sketch the vector diagram according to the relationship. e) v w/g v b/w = 11.11 m/s?(direction unknown) v w/g = 7.50 m/s left v b/g =? (direction straight across) f) t= (800 m/ 8.196 m/s) = 97.6 s v b/g v b/w v b/g = (11.11) 2 (7.50) 2 = 8.20 m/s θ = sin 1 7.50 11.11 = 42.5 0

Some additional points: An object s water speed or air (wind) speed, is by definition, the magnitude of the velocity of the object relative to the water or the air. In the previous example the vector diagrams were convenient right angled triangles. This is a special case. Often one would have to solve using cosine and sine laws, or use components to find the vector in question.