ch3_eclbls

CHAPTER 3 – TWO-DIMENSIONAL MOTION AND VECTORS 3-1 Introduction to Vectors Welcome to vectors!

Vectors indicate direction; scalars do not. Scalar: a quantity that can be completely specified by its magnitude with appropriate units. It has no magnitude but no direction. Vector: physical quantity that has both direction and magnitude.

Vectors can be added graphically. Resultant: a vector representing the sum of two or more vectors

Now let’s talk about vector properties. Vectors can be moved parallel to themselves in a diagram Determining a resultant vector by drawing a vector from the tail of the first vector o the tip of the last vector is known as the TRIANGLE METHOD OF ADDITION. Vector can be added in any order The vector sum of two or more vectors are added, provided that the magnitude and direction of each vector To subtract a vector, add its opposite The negative of a vector is defined as a vector with the same magnitude as the original vector but opposite in direction. Multiplying of dividing vectors by scalars results in vectors.

Equation Station: math \vec{V} _1 - \vec{V} _2 = \vec{V} _1 + (-\vec{V} _2) math math \vec{V} _1 + \vec{V} _2 = \vec{V} _2 + \vec{V} _1 math

3-2 Vector Operations

Use the Pythagorean theorem to find the magnitude of the resultant change in y= the height and change in x= the distance from one edge of he pyramid o the middle o half the pyramid’s width.

Use the tangent function to find the direction of the resultant Vectors should have the magnitude, the length of the resultant, and the angle that it makes.

Resolving vectors into components

Components: the projection of a vector along the axes of a coordinate system Can be divided by a set of perpendicular components. The sine and cosine functions are defined in terms of the lengths of the sides of such right triangles.

Adding Vectors that are not perpendicular Resolving each of the plane’s displacement vectors into their x and y components. Components along each axis be added together and these vector sums will be the two perpendicular components of the resultant.  Equation: math V_r=\sqrt{V__x^2 + V__y^2 } math math tan^-1= V_y/V_x math math \vec{V}_x= Vcos\thetai and \vec{V}_y = sin \theta j math

Practice Problems

a) What distance has the driver traveled? b) What is the driver’s total displacement Answers: a) 23km b) 17km to the east 2. How fast must a truck travel to stay beneath an airplane that is moving 105km/h at an angle of 25 degrees to the ground? Answer: 95km/h
 * 1) A truck driver attempting to deliver some furniture travels 8km east, turns around and travels 3km west, and then travels 12km east to his destination.

3-3 Projectile Motions

Use of components avoids vector multiplication All the kinematic equations from Chapter 2 could be rewritten in terms of vector quantities. Avoid using the complicated vector forms of the equations altogether. Apply the technique of resolving vectors into components. Then you can apply the simpler one-dimensional forms of the equations

Components simplify projectile motion Projectile motion: free-fall with an initial horizontal velocity

Projectiles follow parabolic trajectories

Projectile motion is free fall with an initial horizontal velocity

Use components to analyze objects launched at an angle

Equations:

math \vec{d} =\vec{V}_i \Delta t + \frac{1}{2} \vec{a} \Delta t^2 math math \vec{d}_x = \vec{v}_xi \Delta t + \frac{1}{2} \vec{a}_x \Delta t^2 math math \vec{d}_y = \vec{v}_yi \Delta t + \frac{1}{2} \vec{a}_y \Delta t^2 math math \vec{v}_x = \vec{v}_i cos \thetai = constant math math \vec{v}_yi=\vec{v}_i sin \theta j math math \vec{d}_x = \vec{v}_i cos \theta \Delta ti math math \vec{d}_y = \vec{v}_i sin \theta \Delta t j + \frac{1}{2} \vec{g} \Delta t^2 math math \vec{d}_x= \frac{v^2 sin 2 \theta}{a_g} math

Practice Problems

Answer: 0.66m/s
 * 1) An autographed baseball rolls off a .70 meter high desk and strikes the floor .25 meters away from the base of the desk. How fast was it rolling?

Answers: 2.0s & 4.8m
 * 1) A baseball is thrown at and angle of 25 degrees relative to the ground at a speed of 23.0m/s. If the ball was caught 42.0m from the thrower, how long was it in the air? How high was the tallest spot in the balls path?

WORKS CITED Physics Text Book