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In this chapter we will learn about the kinematics (displacement, velocity, acceleration) of a particle in two or three dimensions. Uniform Circular MotionPowerPoint Presentation

In this chapter we will learn about the kinematics (displacement, velocity, acceleration) of a particle in two or three dimensions. Uniform Circular Motion

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Reading assignment: Chapter 4 (4.4-4.6)

Homework 6: (due Friday, Sept. 9, 2005):

Chapter 4: 31, 43, 47, 55, 64, 71

- In this chapter we will learn about the kinematics (displacement, velocity, acceleration) of a particle in two or three dimensions.
- Uniform Circular Motion
- Superposition principle

This year’s tutors: Jerry Kielbasa, Matt Rave, Christine Carlisle

All sessions will be in room 103 (next to lecture room).

Tutor sessions in semesters past were very successful and received high marks from students.

All students are encouraged to take advantage of this opportunity.

The displacement vector r:

Displacement is the ___________ line between the final and initial position of the particle.

That is the vector _____________ between the final and initial position.

Average velocity v:

Average velocity: Displacement of a particle, Dr, __________ by time interval Dt.

Instantaneous velocity : Limit of the _____________ velocity as Dt approaches zero. The direction v is always _____________ to the particles path.

The instantaneous velocity equals the _____________ of the position vector with respect to time.

The ______________ of the instantaneous velocity vector is called the speed (scalar)

Average acceleration:

Average acceleration: _____________ in the velocity Dv ________ by the time Dt during which the change occurred.

Change can occur in direction and magnitude!

Acceleration points along change in velocity Dv!

Instantaneous acceleration: limiting value of the ratio

as Dt goes to ____________.

Instantaneous acceleration equals the derivative of the velocity vector with respect to ________________.

Two- (or three)-dimensional motion with constant acceleration a

Trick 1:

The equations of motion we derived before (e.g. _________ equations) are still valid, but are now in __________ form.

Trick 2 (Superposition principle):

Vector equations can be broken down into their x- and y- components. Then calculated ____________________.

Position vector:

Velocity vector:

A melon truck brakes right before a ravine and looses a few melons. The melons skit over the edge with an initial velocity of vx = 10.0 m/s.

- Determine the x- and y-components of the velocity at any time and the total velocity at any time.
- Calculate the velocity and the speed of the melon at t = 5.00 s.
- Determine the x- and y-coordinates of the particle at any time t and the position vector r at any time t.
- Graph the path of a melon.

Motion in a circular path at constant speed.

- Velocity is changing, thus there is an acceleration!!
- Acceleration is _______________ to velocity
- Centripetal acceleration is ____________ the center of the circle
- Magnitude of acceleration is
- r is radius of circle The period is:

Moving frame of reference

A boat heading due north crosses a river with a speed of 10.0 km/h. The water in the river has a speed of 5.0 km/h due east.

- Determine the velocity of the boat.
- If the river is 3.0 km wide how long does it take to cross it?

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