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# objectives - PowerPoint PPT Presentation

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.

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### objectives

Exercise: 3.3, 3.5, 3.7

Position, velocity, acceleration vectors

(x2 – x1)i (y2 – y1)j (z2 – z1)k

vav =

+

+

∆t

∆t

∆t

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 =

dt

• 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.

3c

t =

example

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 =

+

+

∆t

∆t

∆t

One dimension

d2x

dvx

dvy

dvz

d2y

d2z

a = i + j + k

a = i + j + k

dt

dt

dt

dt

dt

dt

ay

tanθ =

ax

|a| = √ax2 + ay2 + az2

Instantaneous 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.

• 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.

• 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.

• 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.

Increasing speed

decreasing speed

Example 3.3 The acceleration

• Given:

• Find the parallel and perpendicular components of the instantaneous acceleration at t = 2.0 s

Example 3.4 The acceleration

• 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 The acceleration

• 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.)

3.6 The acceleration

a = 0 The acceleration

objective The acceleration

• Projectile motion

• Homework: Read 3.3; Exercise: 3.9, 3.11, 3.13

• A The acceleration 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 The acceleration

• 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

θ The acceleration

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.

θ The acceleration

Vertical Component

• Acceleration: ay = -g

• Velocity: vy = vy0 – gt = vi0sinθ - gt

• Displacement: y = y0 + vy0∙t - ½ gt2

The magnitude of the position: 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 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

Air resistance isn’t always negligible. projectile is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. The path of the projectile is parabolic.

Example 3.5 projectile is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. The path of the projectile is parabolic.

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 projectile is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. The path of the projectile is parabolic.

• 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.

objective batted baseball

• Projectile

• Homework: 3.15, 3.17, 3.19

Example 3.8 batted baseball

• 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?

• 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).

tan 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).α0 = h / d

Check your understanding 3.3 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).

• 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)?

objective 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).

• Motion in a circle

• Homework: 3.29, 3.31, 3.33

3.4 motion in a circle 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).

• 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

Decreasing speed 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).

Constant speed – uniform circular motion 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).

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.

Uniform circular motion vs. projectile motion the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

• 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.

Example 3.11 the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

Example 3.12 the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

Non uniform circular motion the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

• The acceleration has two components: a (tangent) and a (radial).

Radial component the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

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 the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

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

caution the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

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

• Suppose that the particle experiences 4 times the the instantaneous acceleration at each point is always long a radius of the circle, toward its center. 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.

• Decrease T the instantaneous acceleration at each point is always long a radius of the circle, toward its center.2 by 3 times, or decrease T by √3

3.28

objective the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

• Relative velocity

• Homework: 37, 39, 41

3.5 relative velocity the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

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 one dimension the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

Relative velocity in two or three dimensions the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

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 the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

150 km/h

example the instantaneous acceleration at each point is always long a radius of the circle, toward its center.

• 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?