Physics 103: Lecture 5 2D Motion + Relative Velocities

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# Physics 103: Lecture 5 2D Motion + Relative Velocities - PowerPoint PPT Presentation

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 52D Motion + Relative Velocities
• Today’s lecture will be on
• More on 2D motion

Physics 103, Spring 2004, U.Wisconsin

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

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

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

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

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

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

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

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

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

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

ABC

• 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

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

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

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

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

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

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

Pendulum Motion - 2 Dimensions

Kinematic equations for constant acceleration

DO NOT APPLY for this case.

Physics 103, Spring 2004, U.Wisconsin

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