chapter 3 kinematics in two dimensions; vectors trigonometry review

Chapter 3

Kinematics in Two Dimensions; Vectors

Trigonometry Review

sin

sideadjacent

sideopposite

hypotenuse

sideadjacent

hypotenuse

sideopposite

tan

cos

sin

More Trigonometry

• Pythagorean Theorem

• To find an angle, you need the inverse trig function– for example,

3-2 Addition of Vectors – Graphical Methods

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.

3-2 Addition of Vectors – Graphical MethodsIf 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.


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


Even if the vectors are not at right angles, they can be added graphically by using the “tail-to-tip” method.


The parallelogram method may also be used; here again the vectors must be “tail-to-tip.”

3-3 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 the negative vector:

3-4 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.


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


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


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:

3-5 Projectile Motion

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 θ0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component.

3-7 Projectile Motion Is ParabolicIn 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.

3-6 Solving Problems Involving Projectile Motion

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

3.3 Projectile Motion

Example 3 A Falling Care Package

The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. Determine the time requiredfor the care package to hit the ground.


y ay vfy viy t-1050 m -9.80 m/s2 0 m/s ?


y ay vfy viy t-1050 m -9.80 m/s2 0 m/s ?

221 tatvy yiy 2

21 tay y

s 6.14

sm9.80

m 1050222

ya

yt


Example 4 The Velocity of the Care Package

What are the magnitude and direction of the final velocity ofthe care package?


y ay vfy viy t-1050 m -9.80 m/s2 ? 0 m/s 14.6 s


y ay vfy viy t-1050 m -9.80 m/s2 ? 0 m/s 14.6 s

sm143

s 6.14sm80.90 2

tavv yiyfy


Conceptual Example 5 I Shot a Bullet into the Air...

Suppose you are driving a convertible with the top down.The car is moving to the right at constant velocity. You pointa rifle straight up into the air and fire it. In the absence of airresistance, where would the bullet land – behind you, aheadof you, or in the barrel of the rifle?


Example 6 The Height of a Kickoff

A placekicker kicks a football at and angle of 40.0 degrees andthe initial speed of the ball is 22 m/s. Ignoring air resistance, determine the maximum height that the ball attains.


iv

ixv

iyv

sm1440sinsm22sin iiy vv

sm1740cossm22cos iix vv


y ay vfy viy t? -9.80 m/s2 0 14 m/s


y ay vfy viy t? -9.80 m/s2 0 14 m/s

yavv yiyfy 222 y

iyfy

a

vvy

2

22

m 10

sm8.92

sm1402

2

y


Example 7 The Time of Flight of a Kickoff

What is the time of flight between kickoff and landing?


y ay vfy viy t0 -9.80 m/s2 14 m/s ?


y ay vfy viy t0 -9.80 m/s2 14 m/s ?

221 tatvy yiy

2221 sm80.9sm140 tt

t2sm80.9sm1420

s 9.2t


Example 8 The Range of a Kickoff

Calculate the range R of the projectile.

m 49s 9.2sm17

221

tvtatvx ixxix

Projectile Motion ExampleA canon is fired with a muzzle velocity of 1000 m/s at an angle of 30°. The projectile fired from the canon lands in the water 40 m below the canon. a)What is the range of the projectile.b)What id the velocity of the projectile in x and y when it landsc)What is the landing angle of the projectile

Knownvi = 1000m/sθ = 30°Δy = -40m

UnknownX, vfx, vfy, θ, vf

smsmv

smsmv

tvtatvx

oy

ox

oxxox

/50030sin)/1000(

/86630cos)/1000(

221

30°voy

vix

vi

Projectile Motion ExampleUse y info to find t

221 tatvy yoy

-40m = (500m/s)t + ½(-9.8m/s2)t2

4.9t2 – 500t – 40 = 0 a b c

Solve for t using quadratic equation

t = -(-500) ± (-500)2 – 4(4.9)(-40) 2(-40)

Take the positive root:t = 102.1s

Projectile Motion Example

mssmx

tvx ox

6.884181.102)/866(

smsmv

msmsmv

yavv

yavv

y

y

yoyy

yoyy

/8.500/8.500

)40)(/8.9(2)/500(

2

2

22

2

22

Calculate x:

Calculate vy

Projectile Motion Example

Find landing angle and velocityvx = 866m/s

vf

θvy = -500.8m/s

vf = (866m/s)2 + (-500.8m/s)2 = 1000.4m/s

tan θ = -500.8m/s = -0.578 866m/s

θ = tan-1(-0.578)

θ = -30.04°

1.7 The Components of a Vector

Example

A displacement vector has a magnitude of 175 m and points atan angle of 50.0 degrees relative to the x axis. Find the x and ycomponents of this vector.

rysin

m 1340.50sinm 175sin ry

rxcos

m 1120.50cosm 175cos rx

yxr ˆm 134ˆm 112

A hiker walks 100m north, 130m northeast, and 120m south. Find the displacement vector and the angle measured from the positive x axis.

N

B

C

W A E

S

D θ

Example: Adding Vectors

A30°

B

CR

+x

+yIn an experiment, an object moves along the path described by vectors A, B, C. The resultant of the three vectors is R. The lengthof the three vectors is as follows: A=10m, B=12m, and C=15m.Find the magnitude and direction of R

Rx

RyX Y

A 10mcos 30° = 8.66m

10msin 30° = 5m

B 0 -12m

C -15m 0

R -6.34m -7m

R = (-6.34m)2 + (-7m)2 = 9.44m

θ

tan θ = -7m -6.34m

θ= 47.8° below the negative x axis

chapter 3 kinematics in two dimensions; vectors trigonometry review

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