Kinematics Study of an object’s motion without regard to the forces causing the motion.

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Kinematics Study of an object’s motion without regard to the forces causing the motion.

Transcript of Kinematics Study of an object’s motion without regard to the forces causing the motion.

Kinematics

Study of an object’s motion without regard to the forces causing the motion.

Kinematics asks three questions at a particular instant in time

Where is the object? Position: distance and direction (“displacement”) from a

given reference point. (Ex: 4 meters East of the wall)

How fast is the object moving and in what direction? Instantaneous Velocity: The rate at which the object

changes position. (Ex: 4 meters per second East.)

How fast is the object speeding up, slowing down or changing direction? Acceleration: The rate at which velocity changes. (Ex: 4

meters per second each second East).

Two Types of Quantities in Physics

Scalars: Have “magnitude” (amount with units) only. Ex: Speed: 4 meters per second

Vectors: Have magnitude and direction. Ex: Velocity: 4 meters per second East

Identify the following as a vector or scalar: A mass: 6 kilograms (6 kg) A distance: 6 meters (4m) A displacement: 6 meters (6m) East A position: 6 meters (6m) East of the wall.

Table of Kinematic Quantities

QUESTION SCALAR VECTOR S.I. UNIT

When?

Where?

How Fast?

Changing Velocity?

Table of Kinematic QuantitiesQUESTION SCALAR VECTOR

(vector symbols are in boldface)

S.I. UNIT

When? Time (t) ------------ Second (s)

Where? Distance (d) Position (x, y)Displacement (d)

Meter (m)

How Fast? Instantaneous Speed (v)

Instantaneous Velocity (v)

m/s

Changing Velocity?

Acceleration (a) Acceleration (a) m/s/s, or m/s2

Average Speed and Velocity(Always applies to a TRIP)

• Average Speed = (distance of trip)/(time of trip)

VAV = d/t• Average Velocity = (displacement of trip)/(time of trip)

VAV = d/t(Note: The boldface type means the quantity ia a

vector)

How does average walking pace vary from person to person?

• Each person walks naturally along the taped path around the room’s perimeter. This distance is about equal to 100 feet (assuming each tile is 1 foot square).

• The time for the walker is measured by his/her partner with a stopwatch.

• The average speed is then: vAV = d/t

• Convert from feet per second to miles per hour.

• What is each person’s average velocity?

How do free falling objects differ in acceleration due to gravity?

• Measure the distance (d) from the bottom of the ceiling light fixtures to the floor.

• Drop various balls and time how long it takes each one to hit the floor (t). Describe each ball: Ping-Pong, rubber, large steel, hard pink.

• Calculate the acceleration d = ½ at2

or: a = 2d/t2 for each ball.

Results of Free Falling Ballsd = 2.4 m

Accepted value: g = 9.8 m/s2 = acceleration due to gravity.

Ball Mass (g) Acceleration (a) (m/s2)

Ping-Pong 2.8 9.5

Rubber 8.6 10.0

Hard Pink 9.9 10.0

Large Steel 66.4 10.1

How can we estimate error due to our reaction time?

How High Is This Cliff?1. You’re at the top of a high

cliff, overlooking a river. How can you find out its height?

2. It takes 10 seconds for a dropped rock to hit the river below. How high is the cliff?

3. How Fast is the rock traveling after 3.5 seconds?

4. How Far has the rock traveled after 3.5 seconds?

Case 2: Accelerating from Rest: v0=0, a ≠ 0

A car accelerates from rest at a constant rate of 2 m/s2. Fill out the following table and plot v vs. t and d vs. t

v = a (Δt)

d = ½ a (Δt)2

t (s) v (m/s) d (m)

0 V0=0

1

2

3

4

5

Case 3: Accelerating from an initial velocity: v0≠ 0, a ≠ 0

How can we find equations for v and d as “functions” of time (t)?

•From definition of acceleration, we get: v = v0 + a (Δt)

•From area under v vs. t graph, we get: d = v0 (Δt) + ½ a (Δt)2

Case 3: Accelerating from an initial velocity: v0≠ 0, a ≠ 0

A car accelerates from v0= 4 m/s at a constant

rate of 2 m/s2. Fill out the following table and plot v vs. t and d vs. t

v = v0 + a (Δt)

d = v0 (Δt) + ½ a (Δt)2

t (s) v (m/s) d (m)

0 V0=4

1

2

3

4

5

Summarizing Kinematic CasesΔt = time interval

v0 = velocity at start of time intervalv = velocity at end of time interval

d = distance traveled at end of time interval

• Case 1: Constant velocity: v0 ≠ 0; a = 0

v = v0 , d = v0(Δt)

• Case 2: Accelerating from Rest: v0 = 0; a ≠ 0 v = a (Δt) , d = ½ a (Δt)2

• Case 3: Accelerating from an initial velocity v0 ≠ 0; a ≠ 0 v = v0 + a (Δt) (“Kinematic Equation 1”)

d = v0 (Δt) + ½ a (Δt)2 (“Kinematic Equation 2”)

Summarizing Kinematic Cases(setting initial time at: t0 =0)

• Case 1: Constant velocity: v0 ≠ 0; a = 0

v = v0 , d = v0 t

• Case 2: Accelerating from Rest: v0 = 0; a ≠ 0 v = a t , d = ½ a t2

• Case 3: Accelerating from an initial velocity v0 ≠ 0; a ≠ 0

v = v0 + a t (“Kinematic Equation 1”)

d = v0 t + ½ a t2 (“Kinematic Equation 2”)

Summarizing Kinematic Cases

General Time Interval• Case 1: Constant velocity: v0

≠ 0; a = 0 v = v0 , d = v0(Δt)

• Case 2: Accelerating from Rest: v0 = 0; a ≠ 0 v = a (Δt) , d = ½ a (Δt)2

• Case 3: Accelerating from an initial velocity v0 ≠ 0; a ≠ 0 v = v0 + a (Δt)

(“Kinematic Equation 1”) d = v0 (Δt) + ½ a (Δt)2

(“Kinematic Equation 2”)

Time interval starting at t0 = 0

• Case 1: Constant velocity: v0

≠ 0; a = 0 v = v0 , d = v0 t

• Case 2: Accelerating from Rest: v0 = 0; a ≠ 0 v = a t , d = ½ a t2

• Case 3: Accelerating from an initial velocity v0 ≠ 0; a ≠ 0 v = v0 + a t

(“Kinematic Equation 1”) d = v0 t + ½ a t2

(“Kinematic Equation 2”)

Relative MotionPerson A stands on the sidewalk. Given:•A sees B drive by at 40 mi/h East.•A sees C drive by at 60 mi/h East.Questions:•According to B

What is A’s velocity? What is C’s velocity?

•According to C What is A’s velocity? What is B’s velocity?

•Suppose B throws a ball sideways out the window. How does B see the ball move? How does A see the ball move?