CT1: Suppose you are running at constant velocity along a level track and you wish to throw a ball so you catch it when it comes back down. You should throw the ball with a component of velocity in the forward direction. straight up. with a component of velocity in the backward direction.

Ch4 Motion in Two Dimensions 4.1 The Position, Displacement, Velocity, and Acceleration Vectors

v

CT 2: Which graph represents a constant acceleration? vx vx t t A B

Ch4: Motion in Two Dimensions 4.2: Two Dimensional Motion with Constant Acceleration

Equations for Constant Acceleration Only vxf = vxi + axt xf = xi + (vxi + vxf) t / 2 xf = xi + vxi t + axt2/2 vxf2 = vxi2 + 2ax(xf – xi) Assuming the conditions: ti = 0, tf = t, x(0) = xi and v(0) = vxi and ax is constant.

vyf = vyi + ayt yf = yi + (vyi + vyf) t / 2 yf = yi + vyi t + ayt2/2 vyf2 = vyi2 + 2ay(yf – yi)

P4.5 (p. 93)

Ch4 Motion in Two Dimensions 4.3 Projectile Motion Don’t use the derived projectile motion equations. Start from a drawing, listing knowns and unknowns and the constant acceleration equations.

Concept Question 3

Neglect air resistance Neglect the Earth’s rotation Neglect the small variations in g over the Earth’s surface and with height. CT4: For all the above parabolic trajectories above the time to rise to the maximum height (t1) is related to the time to return to the same level from the maximum height (t2) by t1 = t2 t1 = 2t2 t1 = t2/2 t1 > t2 t1 < t2

Average: 76% Percent Doing: 66% If you got 6 or less, learn vectors!

Before Class Assignment 9/18/2007 q4-1 14/15 correct q4-2 9 correct 1 incorrect 2 answered different question 1 correct explanation but no answer

P4.9 (p. 93) P4.53 (p. 97) P4.18 (p. 94)

CT5: A projectile in freefall has a projection angle less than 90° CT5: A projectile in freefall has a projection angle less than 90°. At what point are the velocity and acceleration parallel? nowhere the highest point the launch point

CT6: A projectile in freefall has a projection angle less than 90° CT6: A projectile in freefall has a projection angle less than 90°. At what point are the velocity and acceleration perpendicular? nowhere the highest point the launch point

The Symmetry of Projectile Motion T/2 - t T/2 + t T The green vectors are velocities.

Figure 4.9  Motion diagram for a projectile. Fig. 4.9, p.85

Active Figure 4.7  The parabolic path of a projectile that leaves the origin with a velocity vi. The velocity vector v changes with time in both magnitude and direction. This change is the result of acceleration in the negative y direction. The x component of velocity remains constant in time because there is no acceleration along the horizontal direction. The y component of velocity is zero at the peak of the path. At the Active Figures link at http://www.pse6.com you can change launch angle and initial speed. You can also observe the changing components of velocity along the trajectory of the projectile. Fig. 4.7, p.84

Ch4 Motion in Two Dimensions 4.4 Uniform Circular Motion Circular path with constant speed.

Concept Question 7 F. E. D. C. B. A.

Figure 4.17  (a) A car moving along a circular path at constant speed experiences uniform circular motion. (b) As a particle moves from A to B, its velocity vector changes from vi to vf. (c) The construction for determining the direction of the change in velocity ∆v, which is toward the center of the circle for small ∆r.

P4.25 (p. 94)

Ch4 Motion in Two Dimensions 4.5 Tangential and Radial Acceleration

r   a = ar + at a = -v2/r r + d|v|/dt  ^ ^ Active Figure 4.18  The motion of a particle along an arbitrary curved path lying in the xy plane. If the velocity vector v (always tangent to the path) changes in direction and magnitude, the components of the acceleration a are a tangential component at and a radial component ar. a = ar + at a = -v2/r r + d|v|/dt  ^ ^

P4.28 (p. 95) P4.31 (p. 95)

Concept Question 8 A. B. C. D.

Concept Questions 9,10,11

Concept Question 12