Skip this Video
Download Presentation

Loading in 2 Seconds...

play fullscreen
1 / 67

objectives - PowerPoint PPT Presentation

  • Uploaded on

objectives. Homework: read 3.1 & 3.2 Exercise: 3.3, 3.5, 3.7. Position, velocity, acceleration vectors. Position vector. (x 2 – x 1 ) i (y 2 – y 1 ) j (z 2 – z 1 ) k. v av =. +. +. ∆t. ∆t. ∆t. Average velocity vector.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

PowerPoint Slideshow about ' objectives' - ryan-page

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript


Homework: read 3.1 & 3.2

Exercise: 3.3, 3.5, 3.7

Position, velocity, acceleration vectors

average velocity vector

(x2 – x1)i (y2 – y1)j (z2 – z1)k

vav =






Average velocity vector
  • During a time interval t the particle moves from P1 , where its position vector is r1, to P2, where its position vector is r2.

One dimension



Instantaneous velocity



v =


test your understanding
Test your understanding
  • In which of these situations would the average velocity vector vav over an interval be equal to the instantaneous velocity v at the end of interval:
  • A body moving along a curved path at constant speed
  • A body moving along a curved path and speeding up
  • A body moving along a straight line at constant speed
  • A body moving along a straight line and speeding up.



t =


If r = bt2i+ ct3j

Where b and c are positive constants, when does the velocity vector make an angle of 45.0o with the x- and y-axes?

v = dr/dt = 2bt i + 3ct2 j

= 1


(v2x – v1x)i (v2y – y1y)j (v2z – v1z)k

aav =






One dimension








a = i + j + k

a = i + j + k








tanθ =


|a| = √ax2 + ay2 + az2

Instantaneous Acceleration


All about acceleration

  • Equal to time rate of change of velocity
  • ≠ 0 if velocity changes in magnitude or direction.
  • It does not have same direction as velocity vector
  • Acceleration vector lies on concave side of curved path.
example 3 2
Example 3.2
  • Given:
  • Find the components of the average acceleration in the interval from t = 0.0 s to t = 2.0 s.
  • Find the instantaneous acceleration at t = 2.0 s, its magnitude and its direction.
parallel and perpendicular components of acceleration
Parallel and perpendicular components of acceleration
  • The acceleration vector a for a particle can describe changes in the particle’s speed, its direction of motion, or both.
  • The component of acceleration parallel to a particle’s path (parallel to the velocity) tells us about changes in the particle\'s speed.
  • The acceleration component perpendicular to the path (perpendicular to the velocity) tells us about changes in the particle’s direction of motion.
the effect of acceleration s directions
The effect of acceleration’s directions
  • When acceleration is perpendicular to particle’s velocity: velocity’s magnitude does not change, only its direction changes, particle moves in a curved path at constant speed.
  • When acceleration is parallel to particle’s velocity: velocity’s magnitude changes only, its direction remains the same, particle moves in a straight line with changing speed.
In most cases, the particle’s speed and direction change. The acceleration a has both parallel and perpendicular components.

Increasing speed

decreasing speed

example 3 3
Example 3.3
  • Given:
  • Find the parallel and perpendicular components of the instantaneous acceleration at t = 2.0 s
example 3 4
Example 3.4
  • A skier moves along a ski-jump ramp as shown in the figure. The rap is straight from point A to point C onward. The skier picks up speed as she moves downhill from point A to point E. draw the direction of the acceleration vector at points B, D, E and F.
test your understanding 3 2
Test your understanding 3.2
  • A sled travels over the crest of a snow-covered hill. The sled slows down as it climbs up one side of the hill and gains speed as it descends on the other side. Which of the vectors (1 though 9) in the figure correctly shows the direction of the sled’s acceleration at the crest? Choice 9 is that the acceleration is zero.)
  • Projectile motion
  • Homework: Read 3.3; Exercise: 3.9, 3.11, 3.13
projectile motion

A projectile is any body that is given an initial velocity and then follows a path determined entirely by the effects of gravitational acceleration and air resistance.

Projectile Motion

projectiles move in two dimensions
Projectiles move in TWO dimensions
  • Since a projectile moves in 2-dimensions, it therefore has 2 components just like a resultant vector.
  • Horizontal and Vertical

The path of a projectile is called a trajectory

horizontal component


Horizontal Component
  • Velocity is constant: vx0 = vi0cosθ
  • Acceleration: ax = 0
  • Displacement x = x0 + vx0∙t

In other words, the horizontal velocity is CONSTANT. BUT WHY?

Gravity DOES NOT work horizontally to increase or decrease the velocity.

vertical component


Vertical Component
  • Acceleration: ay = -g
  • Velocity: vy = vy0 – gt = vi0sinθ - gt
  • Displacement: y = y0 + vy0∙t - ½ gt2
If air resistance is negligible, the trajectory of a projectile is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. The path of the projectile is parabolic.

The magnitude of the position:

The magnitude of the velocity:

The direction of the velocity:

  • If at t = 0, x0 = y0 = 0 then we can find the x, y coordinates and the x, y velocity at time t:

In y direction:

vy = v0sinθ - gt

y = (v0sinθ)t – ½ gt2

vy2 = (v0sinθ)2– 2gy

In x direction:

vx = v0cosθt

x = (v0cosθ)t


Example 3.5

Let’s consider again the skier in example 3.4. What is her acceleration at points G, H, and I after she flies off the ramp? Neglect air resistance.

The acceleration at points G, H, I are the same:

ax = 0; ay = -9.8 m/s2

example 3 6 a body projected horizontally
Example 3.6:a body projected horizontally
  • A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. find the motorcycle’s position,distance and velocity from the edge of the cliff, and velocity after 0.50 s.
  • Projectile
  • Homework: 3.15, 3.17, 3.19
example 3 8
Example 3.8
  • For a projectile launched with speed v0 at initial angle α0 (between 0o and 90o), derive general expressions for the maximum height h and horizontal range R. For a given v0, what value of α0 gives maximum height? What value gives maximum horizontal range?
You toss a ball from your window 8.0 m above the ground. When the ball leaves your hand, it is moving at 10.0 m/s at an angle of 20o below the horizontal. How far horizontally from your window will the ball hit the ground? Ignore air resistance.
A monkey escapes from the zoo and climbs a tree. After failing to entice the monkey down, the zookeeper fires a tranquilizer dart directly at the monkey. The clever monkey lets go at the same instant the dart leaves the gun barrel, intending to land on the ground and escape. Show that the dart always hits the monkey, regardless of the dart’s muzzle velocity (provided that it gets to the monkey before he hits the ground).
check your understanding 3 3
Check your understanding 3.3
  • In example 3.10, suppose the tranquilizer dart has a relatively low muzzle velocity so that the dart reaches a maximum height at a point before striking the monkey. When the dart is at point P, will the monkey be
  • At point A (higher than P)
  • At point B (at the same height as P)
  • At point C (lower than P)?
  • Motion in a circle
  • Homework: 3.29, 3.31, 3.33
3 4 motion in a circle
3.4 motion in a circle
  • When a particle moves along a curved path, the direction of its velocity is tangent to the curve, and its acceleration is pointing toward the concave side.

Increasing speed


There is no component of acceleration parallel (tangent) to the path; the acceleration vector is perpendicular (normal) to the path and hence directed inward toward the center of the circular path.

The subscript “rad” is a reminder that the direction of the instantaneous acceleration at each point is always long a radius of the circle, toward its center.
uniform circular motion vs projectile motion
Uniform circular motion vs. projectile motion
  • the magnitude of acceleration is constant at all times.
  • the direction of the direction of the acceleration in projectile always points down;
  • the magnitude of acceleration is constant at all times.
  • the direction of acceleration changes continuously - always points toward the center of the circle.
non uniform circular motion
Non uniform circular motion
  • The acceleration has two components: a (tangent) and a (radial).

Radial component

arad is always perpendicular to the instantaneous velocity and directed toward the center of the circle. But since v is changing, arad is not constant. arad is greatest at the point in the circle where the speed is greatest.

tangent component
Tangent component

The component of acceleration that is parallel to the instantaneous velocity is the atan because it is tangent to the circle.

atan is equal to the rate of change of speed.

In uniform circular motion, there is no change in speed, atan = 0


The two quantities are not the same.

  • The first, equal to the tangential acceleration, is the
  • rate of change of speed; it is zero whenever a particle moves with constant speed, even when its direction of motion changes.
  • The second, is the magnitude of the vector
  • acceleration; it is zero only when the particle’s acceleration vector is zero – motion in a straight line with constant speed.
    • In uniform circular motion
    • In non uniform circular motion
test your understanding 3 4

Suppose that the particle experiences 4 times the acceleration at the bottom of the loop as it does at the top of loop. Compared to its speed at the top the loop, is its speed the bottom of the loop:

  • √2 times as great
  • 2 times as great
  • 2√2 times as great
  • 4 times as great
  • 16 times as great.
Test your understanding 3.4
  • Relative velocity
  • Homework: 37, 39, 41
3 5 relative velocity
3.5 relative velocity

The velocity seen by a particular observer is called the velocity relative to that observe, or simply relative velocity. What is the planes’ speed?

Relative to each other, the planes are almost at rest

Relative to the observers on the ground, the planes are flying at a great speeds.

relative velocity in two or three dimensions
Relative velocity in two or three dimensions

V(plane to Earth) = V (plane to air) + V (air to Earth)

The acceleration of a(plane to earth) is identical to a(plane to air) because the v(air to earth) is assumed to be constant.


R: 150 km/h

150 km/h

  • A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North.
  • What is the resultant velocity of the motorboat?
  • If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
  • What distance downstream does the boat reach the opposite shore?