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Physics 103: Lecture 5 2D Motion + Relative Velocities

Physics 103: Lecture 5 2D Motion + Relative Velocities. Today’s lecture will be on More on 2D motion Addition of velocities. Summary of Lecture 4 Kinematics in Two Dimensions. x = x 0 + v 0x t + 1/2 a x t 2 v x = v 0x + a x t v x 2 = v 0x 2 + 2a x  x.

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Physics 103: Lecture 5 2D Motion + Relative Velocities

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  1. Physics 103: Lecture 52D Motion + Relative Velocities • Today’s lecture will be on • More on 2D motion • Addition of velocities Physics 103, Spring 2004, U.Wisconsin

  2. Summary of Lecture 4Kinematics in Two Dimensions • x = x0 + v0xt + 1/2 axt2 • vx = v0x +axt • vx2 = v0x2 + 2axx • y = y0 + v0yt + 1/2 ayt2 • vy = v0y +ayt • vy2 = v0y2 + 2ayy x andymotions areindependent! They share a common time t Physics 103, Spring 2004, U.Wisconsin

  3. Question? Without air resistance, an object dropped from a plane flying at constant speed in a straight line will 1. Quickly lag behind the plane. 2. Remain vertically under the plane. 3. Move ahead of the plane There is no acceleration along horizontal - object continues to travel at constant speed (same as that of the plane) along horizontal. Due to gravitational acceleration the object’s speed downwards increases. Physics 103, Spring 2004, U.Wisconsin

  4. correct Lecture 4, Pre-Flight 7&8 A flatbed railroad car is moving along a track at constant velocity. A passenger at the center of the car throws a ball straight up. Neglecting air resistance, where will the ball land? 1. Forward of the center of the car 2. At the center of the car 3. Backward of the center of the car Physics 103, Spring 2004, U.Wisconsin

  5. correct Lecture 5, Pre-Flight 1&2 You are a vet trying to shoot a tranquilizer dart into a monkey hanging from a branch in a distant tree. You know that the monkey is very nervous, and will let go of the branch and start to fall as soon as your gun goes off. On the other hand, you also know that the dart will not travel in a straight line, but rather in a parabolic path like any other projectile. In order to hit the monkey with the dart, where should you point the gun before shooting? 1 Right at the monkey 2 Below the monkey 3 Above the monkey If the shot is fired at the monkey the same time the monkey drops, both objects will fall at the same rate causing the shot to hit the monkey. since the monkey is going to start falling right away you need to aim below it Along the way, gravity is going to pull the dart down, so you would have to aim up. Aiming right at it or below would miss the monkey by going underneath it. Physics 103, Spring 2004, U.Wisconsin

  6. Dart hits the monkey! Shooting the Monkey... x= v0 t y= -1/2 g t2 x = x0 y= -1/2 g t2 Physics 103, Spring 2004, U.Wisconsin

  7. Dart hits the monkey! Shooting the Monkey... r= r0 - 1/2 g t2 • At an angle, still aim at the monkey! r= v0t - 1/2 g t2 Physics 103, Spring 2004, U.Wisconsin

  8. Shooting the Enemy Paratrooper... If a soldier wants to shoot down an enemy paratrooper descending at uniform speed, Se, where should he aim? • Above the enemy • At the enemy • Below the enemy • Answer depends on the enemy’s position and vertical speed. r= r0 - Se t r= v0t - 1/2 g t2 Miss the enemy Physics 103, Spring 2004, U.Wisconsin

  9. Reference Frames: Relative Motion VCB=VCA+VAB Velocity of B relative to ground ( C ) : VCB Velocity of A relative to ground ( C ) : VCA Velocity of B relative to A : VAB Physics 103, Spring 2004, U.Wisconsin

  10. Relative Motion • If an airplane flies in a jet stream, depending on the relative orientation of the airplane and the jet stream, the plane can go faster or slower than it normally would in the absence of the jet stream • If a person rows a boat across a rapidly flowing river and tries to head directly for the shore, the boat moves diagonally relative to the shore • Velocity is a vector - add velocities like vectors • Sum the components • Vx = V1x + V2x • V1x = V1 cosq • Vy = V1y + V2y • V1x = V1 sinq Physics 103, Spring 2004, U.Wisconsin

  11. correct ABC Lecture 5, Pre-Flight 3&4 Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a river in the least time. Relative to the water, Beth (B) swims perpendicular to the flow, Ann (A) swims upstream, and Carly (C) swims downstream. Which swimmer wins the race? A) Ann B) Beth C) Carly Physics 103, Spring 2004, U.Wisconsin

  12. ABC Lecture 5, Pre-Flight 3&4(great answers) • Beth will reach the shore first because the vertical component of her velocity is greater than that of the other swimmers. • The key here is how fast the vector in the vertical direction is. "B" focuses all of its speed on the vertical vector, while the others divert some of their speed to the horizontal vectors. Time to get across = width of river/vertical component of velocity Physics 103, Spring 2004, U.Wisconsin

  13. ABC Lecture 5, Pre-Flight 3&4(common misconceptions) • While Carly is moving forward she will also be moving along with the current. two positive(+) direction motions = faster velocity. • Carly will get there first because she is using the current to her advantage. Physics 103, Spring 2004, U.Wisconsin

  14. correct vwg vsg vsw Followup Question Heather wants to swim across a flowing river in such a way that she ends up on the opposite side directly opposite her starting point. She should therefore aim…. 1) upstream 2) downstream 3) directly across Physics 103, Spring 2004, U.Wisconsin

  15. correct Lecture 5, Pre-Flight 5 and 6 A seagull flies through the air with a velocity of 9 m/s if there were no wind. However, it is making the same effort and flying in a headwind. If it takes the bird 20 minutes to travel 6 km as measured on the earth, what is the velocity of the wind? 1. 4 m/s 2. -4 m/s 3. 13 m/s 4. -13 m/s • Seagull’s velocity in the reference frame of the wind = 9 m/s • i. e., in this frame, wind velocity is zero • Seagull travels at 6000/1200 = 5 m/s relative to earth. Therefore, the wind velocity relative to earth is 5-9=-4 m/s. Physics 103, Spring 2004, U.Wisconsin

  16. correct Follow-up, Pre-Flight 5 and 6 If the seagull turns around and flies back how long will it take to return? 1. More time than for flying out 2. Less time than for flying out 3. The same amount of time • Seagull’s return velocity is: -4-9=-13 m/s. The speed is higher so it takes less time to return. Time taken for the return is given by 6000 m / 13 (m/s) = 461.5 s = 461.5/60 = 7.69 minutes Physics 103, Spring 2004, U.Wisconsin

  17. correct Follow-up 2, Pre-Flight 5 and 6 How are the round-trip times with and without wind related if the seagull always goes at 9 m/s? 1. The round-trip time is the same with/without the wind 2. The round trip time is always larger with the wind 3. It is not possible to calculate this • Time taken for the round trip with wind is: 27.69 minutes • Time taken for the round trip without wind is: 12000 m / 9 m/s = 1333 s = 22.2 minutes Physics 103, Spring 2004, U.Wisconsin

  18. Once again: correct A boat is drifting in a river which has a current of 1 mph. The boat is a half mile upstream of a rock when an observer on the boat sees a seagull overhead. The observer sees the gull flying continuously back and forth at constant speed (10 mph) between the boat and the rock. When the boat passes close by the rock, how far (what distance) has the gull flown? • 5 miles • 10 miles • Not sufficient information to determine the distance Physics 103, Spring 2004, U.Wisconsin

  19. Pendulum Motion - 2 Dimensions Kinematic equations for constant acceleration DO NOT APPLY for this case. Physics 103, Spring 2004, U.Wisconsin

  20. Summary Relative Motion Velocity of seagull with respect to ground, Vsg Velocity of seagull with respect to wind, Vsw(i.e., velocity with which it would fly in calm winds) Velocity of wind with respect to the ground, Vwg Physics 103, Spring 2004, U.Wisconsin

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