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Chapter 3 Kinematics in Two Dimensions; Vectors

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Trigonometry Review

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More Trigonometry Pythagorean Theorem To find an angle, you need the inverse trig function –for example,

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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.

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3-2 Addition of Vectors – Graphical Methods 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.

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3-2 Addition of Vectors – Graphical Methods Adding the vectors in the opposite order gives the same result:

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3-2 Addition of Vectors – Graphical Methods Even if the vectors are not at right angles, they can be added graphically by using the “tail-to-tip” method.

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3-2 Addition of Vectors – Graphical Methods The parallelogram method may also be used; here again the vectors must be “tail-to-tip.”

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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:

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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.

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3-4 Adding Vectors by Components If the components are perpendicular, they can be found using trigonometric functions.

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3-4 Adding Vectors by Components The components are effectively one-dimensional, so they can be added arithmetically:

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3-4 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:

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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.

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It can be understood by analyzing the horizontal and vertical motions separately. 3-5 Projectile Motion

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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.

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3-5 Projectile Motion 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.

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3-7 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.

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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.

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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 required for the care package to hit the ground.

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3.3 Projectile Motion yayay v fy v iy t -1050 m-9.80 m/s 2 0 m/s?

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3.3 Projectile Motion yayay v fy v iy t -1050 m-9.80 m/s 2 0 m/s?

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3.3 Projectile Motion Example 4 The Velocity of the Care Package What are the magnitude and direction of the final velocity of the care package?

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3.3 Projectile Motion yayay v fy v iy t -1050 m-9.80 m/s 2 ?0 m/s14.6 s

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3.3 Projectile Motion yayay v fy v iy t -1050 m-9.80 m/s 2 ?0 m/s14.6 s

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3.3 Projectile Motion 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 point a rifle straight up into the air and fire it. In the absence of air resistance, where would the bullet land – behind you, ahead of you, or in the barrel of the rifle?

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3.3 Projectile Motion Example 6 The Height of a Kickoff A placekicker kicks a football at and angle of 40.0 degrees and the initial speed of the ball is 22 m/s. Ignoring air resistance, determine the maximum height that the ball attains.

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3.3 Projectile Motion

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yayay v fy v iy t ?-9.80 m/s 2 014 m/s

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3.3 Projectile Motion yayay v fy v iy t ?-9.80 m/s 2 014 m/s

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3.3 Projectile Motion Example 7 The Time of Flight of a Kickoff What is the time of flight between kickoff and landing?

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3.3 Projectile Motion yayay v fy v iy t 0-9.80 m/s 2 14 m/s?

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3.3 Projectile Motion yayay v fy v iy t 0-9.80 m/s 2 14 m/s?

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3.3 Projectile Motion Example 8 The Range of a Kickoff Calculate the range R of the projectile.

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Projectile Motion Example A 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 lands c)What is the landing angle of the projectile Known v i = 1000m/s θ = 30° Δy = -40m Unknown X, v fx, v fy, θ, v f 30° v oy v ix vivi

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Projectile Motion Example Use y info to find t -40m = (500m/s)t + ½(-9.8m/s 2 )t 2 4.9t 2 – 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

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Projectile Motion Example Calculate x: Calculate v y

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Projectile Motion Example Find landing angle and velocity v x = 866m/s vfvf θ v y = -500.8m/s v f = (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°

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Projectile Motion Example Find landing angle and velocity v x = 866m/s vfvf θ v y = -500.8m/s v f = (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°

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1.7 The Components of a Vector Example A displacement vector has a magnitude of 175 m and points at an angle of 50.0 degrees relative to the x axis. Find the x and y components of this vector.

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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

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Example: Adding Vectors A 30° B C R +x +y In an experiment, an object moves along the path described by vectors A, B, C. The resultant of the three vectors is R. The length of the three vectors is as follows: A=10m, B=12m, and C=15m. Find the magnitude and direction of R Rx Ry XY A10mcos 30° = 8.66m 10msin 30° = 5m B0-12m C-15m0 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

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