Presentation is loading. Please wait.

Presentation is loading. Please wait.

Physics 215 – Fall 2014Lecture 03-11 Welcome back to Physics 215 Today’s agenda: Velocity and acceleration in two-dimensional motion Motion under gravity.

Similar presentations


Presentation on theme: "Physics 215 – Fall 2014Lecture 03-11 Welcome back to Physics 215 Today’s agenda: Velocity and acceleration in two-dimensional motion Motion under gravity."— Presentation transcript:

1 Physics 215 – Fall 2014Lecture Welcome back to Physics 215 Today’s agenda: Velocity and acceleration in two-dimensional motion Motion under gravity -- projectile motion Acceleration on curved path

2 Physics 215 – Fall 2014Lecture Current homework assignment HW2: –Ch.2 (Knight textbook): 48, 54 –Ch.3 (Knight textbook): 28, 34, 40 –Ch.4 (Knight textbook): 42 –due Wednesday, Sept 10 th in recitation

3 Physics 215 – Fall 2014Lecture Exam 1: next Thursday (9/18/14) In room 208 (here!) at the usual lecture time Material covered: –Textbook chapters –Lectures up through 9/16 (slides online) –Wed/Fri Workshop activities –Homework assignments Work through practice exam problems (posted on Blackboard) Work on more practice exam problems next Wednesday in recitation workshop

4 Physics 215 – Fall 2014Lecture Motion under gravity a y = -g v y = v 0y - gt y = y 0 + v 0y t - (1/2)gt 2 a x = 0 v x = v 0x x = x 0 + v 0x t x y v0v0  v 0y = v 0 sin(  ) v 0x = v 0 cos(  ) Projectile motion...

5 Physics 215 – Fall 2014Lecture Projectile question A ball is thrown at 45 o to vertical with a speed of 7 m/s. Assuming g=10 m/s 2, how far away does the ball land?

6 Physics 215 – Fall 2014Lecture A battleship simultaneously fires two shells at enemy ships. If the shells follow the parabolic trajectories shown, which ship will be hit first? A. A B. Both at the same time C. B D. need more information

7 Physics 215 – Fall 2014Lecture Projectile motion R : when is y=0 ? t[v y1 -(1/2)gt] = 0 i.e., T = (2v)sin  /g  R (x-eqn.) max (y-eqn.)  h max (y-eqn.) y x

8 Physics 215 – Fall 2014Lecture Maximum height and range

9 Physics 215 – Fall 2014Lecture Motion on a curved path at constant speed Is the acceleration of the object equal to zero?

10 Physics 215 – Fall 2014Lecture Velocity is tangent to path O sIsI sFsF ss v =  s/  t lies along dotted line. As  t  0 direction of v is tangent to path

11 Physics 215 – Fall 2014Lecture Motion on a curved path at constant speed

12 Physics 215 – Fall 2014Lecture Subtracting vectors vFvF -vI-vI vv same as vFvF vIvI Recall that v F + (-v I ) =  v vv

13 Physics 215 – Fall 2014Lecture For an object moving at constant speed along a curved path, the acceleration is not zero.

14 Physics 215 – Fall 2014Lecture For which of the following motions of a car does the change in velocity vector have the greatest magnitude? (All motions occur at the same constant speed.) A.A 90° right turn at constant speed B.A U-turn at constant speed C.A 270° turn on a highway on-ramp D.The change in velocity is zero for all three motions.

15 Physics 215 – Fall 2014Lecture A. B. C.D. A car moves along the path shown. Velocity vectors at two different points are sketched. Which of the arrows below most closely represents the direction of the average acceleration?

16 Physics 215 – Fall 2014Lecture A. B. C.D. A child is riding a bicycle on a level street. The velocity and acceleration vectors of the child at a given time are shown. Which of the following velocity vectors may represent the velocity at a later time? a

17 Physics 215 – Fall 2014Lecture A biker is riding at constant speed clockwise on the oval track shown below. Which vector correctly describes the acceleration at the point indicated?

18 Physics 215 – Fall 2014Lecture Biker moving around oval at constant speed As point D is moved closer to C, angle approaches 90°.

19 Physics 215 – Fall 2014Lecture Summary For motion at constant speed, instantaneous acceleration vector is perpendicular to velocity vector Points ``inward’’ What is the magnitude of the acceleration vector?

20 Physics 215 – Fall 2014Lecture Acceleration vectors for ball swung in a horizontal circle at constant speed v What is the magnitude of the acceleration?  a  = v 2 /R   R v1v1 v2v2 v1v1 v2v2

21 Physics 215 – Fall 2014Lecture Acceleration of object moving at constant speed on a circular path: Acceleration depends on radius of circle.

22 Physics 215 – Fall 2014Lecture Two cars are moving at different constant speeds on a curved road. One after the other, they are passing the same point on the road: Car A at 18 mph; car B at 36 mph. If car A’s acceleration is 2 m/s 2, car B’s acceleration is: A.1 m/s 2 B.2 m/s 2 C.4 m/s 2 D.8 m/s 2

23 Physics 215 – Fall 2014Lecture What is the magnitude of the acceleration of an object moving at constant speed if the path is curved but not a circle? “r” is the radius of curvature of the path at a given point

24 Physics 215 – Fall 2014Lecture Acceleration vector for object speeding up from rest at point A ?

25 Physics 215 – Fall 2014Lecture What if the speed is changing? Consider acceleration for object on curved path starting from rest Initially, v 2 /r = 0, so no radial acceleration But a is not zero! It must be parallel to velocity

26 Physics 215 – Fall 2014Lecture Acceleration vectors for object speeding up: Tangential and radial components (or parallel and perpendicular)

27 Physics 215 – Fall 2014Lecture Reading assignment Circular motion 4.5 – 4.7 in textbook


Download ppt "Physics 215 – Fall 2014Lecture 03-11 Welcome back to Physics 215 Today’s agenda: Velocity and acceleration in two-dimensional motion Motion under gravity."

Similar presentations


Ads by Google