2.1 Linear Motion

•Displacement

•Velocity

•Average velocity is defined as the change in displacement over the time, t

•Instantaneous velocity is called the limit of x/t as approaches zero or the derivative for x with respect to time, t

  • Direction of v is a tangent to the path at that point in the direction of motion
  • Magnitude of v is the speed

•Acceleration

•Average acceleration define as the ratio of the change in the velocity V=V2-V1 to the time interval t=t2-t1

•Instantaneous acceleration limit of V/t as approaches zero or the derivative for x with respect to time, t

•Magnitude change of v only - 1D motion

•Direction change of v only - circular motion

•Uniform motion

2.2 Motion with Constant a

A constant acceleration means that the acceleration does not depend on time:

Integrating this equation, the velocity of the object can be obtained

where v0 is the velocity of the object at time t = 0.

From the velocity, the position of the object as

function of time can be calculated:

where x0 is the position of the object at time t = 0

Graph of velocity vs time with a constant

Displacement = graph area

graph gradient = acceleration

velocity at any later time t,

Displacement s, at any later time t,

And

2.3 Free falling motion

What you understand about freely falling bodies ?

Free Fall Equation

The Equation of Motion for the 1-D motion of an object in free fall near the Earth’s surface are obtained from equations

If we use the coordinate system shown,then,when substituting into equation

a = g = -9.8 ms-2

Since g points down

This is a nice example of constant acceleration (gravity):

In this case, acceleration is caused by the force of gravity:

Usually pick y-axis “upward”

Acceleration of gravity is “down”:

Problem 2.1.1

•Stone was thrown straight upward at t=0 with +20.0m/s initial velocity on the roof of a 50.0m high building,

–Find the time the stone reaches at maximum height (v=0)

–Find the maximum height

–Find the time the stone reaches its original height

–Find the velocity of the stone when it reaches its original height

–Find the velocity and position of the stone at t=5.00s

1.

2.

3.

4.

Problem 2.1.2

•The pilot of a hovering helicopter drops a lead brick from a height of 1000 m. How long does it take to reach the ground and how fast is it moving when it gets there? (neglect air resistance)

Solution:

•First choose coordinate system.

–Origin and y-direction.

•Next write down position equation:

–

•Realize that v0y = 0.

•Solve for time t when y = 0 given that y0 = 1000 m.

–

•Recall:

•Solve for vy:

Projectile Motion

•A 2-dim motion of an object under the gravitational acceleration with the assumptions

–Free fall acceleration, -g, is constant over the range of the motion

–Air resistance and other effects are negligible

–

•A motion under constant acceleration!!!!  Superposition of two motions

Horizontal motion with constant velocity and Vertical motion under constant acceleration

Show that a projectile motion is a parabola!!!

Example 4.1

A ball is thrown with an initial velocity v=(20i+40j)m/s. Estimate the time of flight and the distance the ball is from the original position when landed.

Which component determines the flight time and the distance?

HorizontalRange and Max Height

•Based on what we have learned previously, one can analyze a projectile motion in more detail

–Maximum height an object can reach

–Maximum range

MaximumRange and Height

•What are the conditions that give maximum height and range in a projectile motion?

Example 4.2

•A stone was thrown upward from the top of a building at an angle of 30o to horizontal with initial speed of 20.0m/s. If the height of the building is 45.0m, how long is it before the stone hits the ground?

What is the speed of the stone just before it hits the ground?