Warm-Up 09/02/10 Vera is speeding down the interstate at 45.0 m/s when she sees an accident in the middle of the road. By the time Vera slams on the breaks, she is 50.0 m from the pileup and slows down at a rate of m/s 2. 1) Construct a velocity-time plot for Vera Side's motion. Use the plot to determine the distance which Vera would travel prior to reaching a complete stop (if she did not collide with the pileup). 2) Use kinematic equations to determine the distance which Vera Side would travel prior to reaching a complete stop (if she did not collide with the pileup). 3) Will Vera hit the cars in the pileup?

The distance traveled can be found by a calculation of the area between the line on the graph and the time axis. Area = ½ b*h = 0.5(4.5 s)(45.0 m/s) Area = 101 m

Given: v i = 45.0 m/s v f = 0.0 m/s a = m/s 2 Find: d = (x f -x i ) =?? v f 2 = v i 2 + 2a (x f -x i ) (0 m/s) 2 = (45.0 m/s) 2 + 2(-10.0 m/s 2 )d ( m 2 /s 2 )/(-20.0 m/s 2 ) =d 101 m =d Since the accident pileup is less than 101 m from Vera, she will indeed hit the pileup before completely stopping (unless she veers aside).

Components of Motion 1. Motion in two dimensions, or curvilinear motion, is motion in which an object moves in a plane that can be described by a rectangular coordinate system. 2. Motion can be resolved, or analyzed, into rectangular components.

An object in motion on a plane can be located using two numbers—the x and y coordinates of its position. Similarly, its velocity can be described using components along the x- and y-axes.

For the ball rolling along the table: The velocity components are: The magnitude of the velocity vector is:

Vector Components – Can describe displacement, velocity, acceleration, and any other vector quantity.

Example 1  An airplane is moving at a velocity of 250 mi/h in a direction 35 o north of east. Find the components of the plane’s velocity in the eastward and northward direction. Given v = 250 mi/h  = 35 o Find V east and V north

SOLUTION V east = V cos  = (250 mi/h)cos35 o = 205 mi/h V north = V sin  = (250 mi/h)sin35 o = 143 mi/h

Displacement Components and Kinematic Calculations The components of the displacement are then given by: Note that the x- and y-components are calculated separately.

More Equations of Motion The equations of motion are: When solving two-dimensional motion problems, each component, x and y, is treated separately. The time is common to both.

If a motor boat were to head straight across a river (that is, if the boat were to point its bow straight towards the other side), it would not reach the shore directly across from its starting point. The river current influences the motion of the boat and carries it downstream.

Example 2  A boat travels with a speed of 5.0 m/s in a straight path on a still lake. Suddenly, a steady wind pushed the boat perpendicular to its straight-line path with a speed of 3.0 m/s for 5.0 s. Relative to the position just when the wind started to blow, where is the boat at the end of this time?  Step One: Sketch the Problem ? 3 m/s  =? 5 m/s

Solution: Choose the x-axis as the original direction of the boat and y-axis as the direction of the wind. Given: V xo = 5.0 m/sa x = 0 m/s 2 V yo = 3.0 m/s t = 5.0 sa y = 0 m/s 2 Find x and y x = x 0 + V xo t y = y 0 + V yo t

Solve for x and y x = 0 + (5.0 m/s)(5s) = 25 m y = 0 + (3.0 m/s)(5s) = 15 m d = sqrt(x 2 + y 2 ) = sqrt((25m) 2 +(15m) 2 )= 29m And tanq = y/x  = tan -1 (y/x) = tan -1 (15m/25m) = 31 o So the boat is now 29 m from where it started, at 31 o to the horizontal.

Curvilinear Motion If the acceleration is not parallel to the velocity, the object will move in a curve:

The diagram below shows the trajectory for a projectile launched non-horizontally from an elevated position on top of a cliff. The initial horizontal and vertical components of the velocity are 8 m/s and 19.6 m/s respectively. Positions of the object at 1-second intervals are shown. Determine the horizontal and vertical velocities at each instant shown in the diagram.

Physlet Exploration 3.1: Addition of Displacement Vectors Δ X 0-8 = = -40, Δy 0-8 = = -16 R 0-8 = 43, q = 21 o N of x axis **movement in the positive x and y direction, change signs to reflect this movement Δ X 8-16 = 18-2 = 16, Δy 8-16 = 4-12 = -8 R 8-16 = -18, q = 26 o N of x-component ** movement in the negative x and positive y directions, change signs to reflect this movement Δx 0-16 = = 24, Δy 0-16 = 16+8 = 24 R 0-16 = 34, q = 45 o N of x axis

Warm-Up 09/07/10 1. A plane flies with a velocity of 52 m/s east through a 12 m/s cross wind blowing the plane south. Find the magnitude and direction (relative to due east) of the resultant velocity at which it travels. 2. An ambitious hiker walks 25 km west and then 35 km south in a day. Find the magnitude and direction (relative to due west) of her resultant displacement.

1. A plane intends to fly north with a speed of 250 m/s relative to the ground through a high altitude cross wind of 50 m/s coming from the east. Determine… A. The bearing that the plane should take (relative to due north and B. The plane’s speed with respect to the air.