velocity, acceleration, free fall ch4 reg
TRANSCRIPT
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Constant VelocityChapter 4Section 4.1-4.3 and 4.7
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What do you think?
Is the book on your instructor’s desk in motion?Explain your answer.
What are some examples of motion?
How would you define motion?
+One-Dimensional Motion
It is the simplest form of motion
This is motion that happens in one direction
Example:Train on the tracts moving forward or
backward
Displacement (x)Change in position
Can be positive or negativeIt describes the direction of motion
You will see me use d instead of xfor displacement
+Displacement (x)
Displacement is not always equal to the distance traveled
A gecko runs up a tree from the 20 cm marker to the 80 cm marker, then he retreats to the 50 cm marker?
+Displacement (x)
In total it traveled 90 cm, however, the displacement is only 30 cm
x or d= 50cm- 20cm = 30 cm
If the gecko goes back to the startThe displacement is zero
Displacement
What is the displacement for the objects shown?
Answer: 70 cm
• Answer: -60 cm
+Average Velocity
Basically velocity is distance/time
• Average velocity is displacement divided by the time interval. d df-di
+Average Velocity
The units can be determined from the equation.SI Units: m/sOther Possible Units: mi/h, km/h,
cm/year
Velocity can be positive or negative but time must always be positive
+Average Velocity
You travel 370 km west to a friends house. You left at 10am and arrived at 3pm. What was your average velocity?
Vavg = d = -370 km
Δt 5.0h
Vavg= -74 km/h or 74 km/h west
+Classroom Practice Problems
Joey rides his bike for 15 min with an average velocity of 12.5 km/h, how far did he ride?
Vavg = d
Δt
d = (12.5 km/h) (0.25h)
d = 3.125 km
+Speed
Speed does not include direction while velocity does.
Speed uses distance rather than displacement.
In a round trip, the average velocity is zero but the average speed is not zero.
+Graphs of Motion
On a speed-versus-time graph the slope represents speed per time, or acceleration.
+Graphs of Motion
Equations and tables are not the only way to describe relationships such as velocity and acceleration.
Graphs can visually describe relationships.
+Graphs of Motion
Speed-Versus-Time On a speed-versus-
time graph, the speed v of a freely falling object can be plotted on the vertical axis and time t on the horizontal axis.
+Graphs of Motion
This particular linearity is called a direct proportion, and we say that time and speed are directly proportional to each other.
+Graphs of Motion
The curve is a straight line, so its slope is constant.
Slope is the vertical change divided by the horizontal change for any part of the line.
+Graphs of Motion
Distance-Versus-TimeWhen the distance d traveled by a
freely falling object is plotted on the vertical axis and time t on the horizontal axis, the result is a curved line.
+Graphs of Motion
This distance-versus-time graph is parabolic.
+Graphs of Motion
The relationship between distance and time is nonlinear.
The relationship is quadratic and the curve is parabolic—when we double t, we do not double d; we quadruple it. Distance depends on time squared!
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AccelerationChapter 4Sections 4.4- 4.6
+What do you think?
Which of the following cars is accelerating?A car shortly after a stoplight turns greenA car approaching a red lightA car with the cruise control set at 80 km/hA car turning a curve at a constant speed
Based on your answers, what is your definition of acceleration?
Acceleration
What are the units? SI Units: m/s2
Other Units: (km/h)/s or (mi/h)/s
Acceleration = 0 implies a constant velocity (or rest)
+Acceleration
In physics, the term acceleration applies to decreases as well as increases in speed.
The brakes of a car can produce produce a large decrease per second in the speed. This is often called deceleration.
+Acceleration
A car is accelerating whenever there is a change in its state of motion.
+Acceleration
A car is accelerating whenever there is a change in its state of motion.
+Acceleration
A car is accelerating whenever there is a change in its state of motion.
+Classroom Practice Problem
A bus slows down with an average acceleration of -1.8m/s2. how long does it take the bus to slow from 9.0m/s to a complete stop?
Find the acceleration of an amusement park ride that falls from rest to a velocity of 28 m/s downward in 3.0 s.9.3 m/s2 downward
+Acceleration
Displacement with constant acceleration
d = ½ (vi + vf) Δt
Displacement= ½ (initial velocity + final velocity) (time)
+Example
A racecar reaches a speed of 42m/s. it uses its parachute and breaks to stop 5.5s later. Find the distance that the car travels during breaking.
Given: vi = 42 m/s vf= 0 m/s
t = 5.5 sec d = ??
+Example
d = ½ (vi + vf) Δt
d = ½ (42+0) (5.5)
d = 115.5 m
+Acceleration
Final velocityDepends on initial velocity,
acceleration and time
Velocity with constant acceleration
vf = vi + aΔt
Final velocity = initial velocity + (acceleration X time)
+Acceleration
Displacement with constant acceleration
d = viΔt + ½ a (Δt)2
Displacement = (initial velocity X time) + ½ acceleration X time2
+Example
A plane starting at rest at one end of a runway undergoes uniform acceleration of 4.8 m/s2 for 15s before takeoff. What is its speed at takeoff? How long must the runway be for the plane to be able to take off?
Given: vi = 0 m/s a= 4.8m/ss t= 15s
vf = ?? d= ??
+Example
What do we need to figure out?Speed at takeoffHow long the runway needs to be
vf = vi + aΔt
vf = 0m/s + (4.8 m/s2)(15 s)
vf = 72m/s
+Example
d = viΔt + ½ a (Δt)2
d = (0m/s)(15s) + ½ (4.8m/s2)(15s)2
d = 540 m
+Acceleration
Final velocity after any displacement
vf2 = vi
2 +2ad
Final velocity2 = initial velocity2 + 2(acceleration)(displacement)
+Example
An aircraft has a landing speed of 83.9 m/s. The landing area of an aircraft carrier is 195 m long. What is the minimum uniform acceleration required for safe landing?
+Example
vf2 = vi
2 +2ad
02= (83.9)2 + 2a (195)
0= 7039.21 + 390a
-7039.21 = 390a
a = -18.0 m/s2
+Classroom Practice Problems
A bicyclist accelerates from 5.0 m/s to 16 m/s in 8.0 s. Assuming uniform acceleration, what distance does the bicyclist travel during this time interval?Answer: 84 m
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2.3 Falling Objectsand section 6.6
+Free Fall
The motion of a body when only the force due to gravity is acting
Acceleration is constant for the entire fall
Acceleration due to gravity (ag or g)
Has a value of -9.81 m/s2 (we will use 10m/s2)Negative for downward
Roughly equivalent to -22 (mi/h)/s
+Free Fall
In Galileo’s famous demonstration, a 10-kg cannonball and a 1-kg stone strike the ground at practically the same time.
+6.6 Free Fall Explained
The ratio of weight (F) to mass (m) is the same for the 10-kg cannonball and the 1-kg stone.
+6.6 Free Fall Explained
The weight of a 1-kg stone is 10 N at Earth’s surface. Using Newton’s second law, the acceleration of the stone is
The weight of a 10-kg cannonball is 100 N at Earth’s surface and the acceleration of the cannonball is
+Free Fall
Acceleration is constant during upward and downward motion
Objects thrown into the air have a downward acceleration as soon as they are released.
+Free Fall
The instant the velocity of the ball is equal to 0 m/s is the instant the ball reaches the peak of its upward motion and is about to begin moving downward.
REMEMBER!!! Although the velocity is 0m/s the
acceleration is still equal to 10 m/s2
+Example
Jason hits a volleyball so that it moves with an initial velocity of 6m/s straight upward. If the volleyball starts from 2.0m above the floor how long will it be in the air before it strikes the floor?
We want to use our velocity and acceleration equations
+Example
Given: vi = 6m/s a= 10m/s2
d= -2.0m t = ?? vf = ??
vf2= vi
2 + 2ad
vf= at + vi
+Free Fall
For a ball tossed upward, make predictions for the sign of the velocity and acceleration to complete the chart.
Velocity (+, -, or zero)
Acceleration(+, -, or zero)
When halfway up
When at the peak
When halfway down
+ -
zero -
- -
Graphing Free Fall
Based on your present understanding of free fall, sketch a velocity-time graph for a ball that is tossed upward (assuming no air resistance). Is it a straight line? If so, what is the slope?
Compare your predictions to the graph to the right.