1 – D Motion

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# 1 D Motion - PowerPoint PPT Presentation

1 – D Motion. We begin our discussion of kinematics (description of motion in mechanics) Simplest case: motion of a particle in 1 – D Concept of a particle : idealization of treating a body as a single point (we get close to doing this by pinpointing the “license plate of a car,” etc.)

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Presentation Transcript
1 – D Motion
• We begin our discussion of kinematics (description of motion in mechanics)
• Simplest case: motion of a particle in 1 – D
• Concept of a particle: idealization of treating a body as a single point (we get close to doing this by pinpointing the “license plate of a car,” etc.)
• 1 – D motion: Only two possible directions
• We will be working with vector quantities
• Characterized by both a magnitude AND a direction
• Scalar quantities have a magnitude but NOT a direction
• How do we describe motion in physics?
• Should be able to answer questions, “Where is it?” along with “Where is it going?” and “How fast is it going?”

+x

Position vectors (keep track of position of car)

Displacement vector (change in position from point 1 to point 2)

Position and Displacement
• Example: Motion of a car along Sandusky St.

(north)

Park Ave. (reference point or origin)

William St.

(point 1)

Central Ave. (point 2)

CQ1: Consider the figure below that shows three paths between position 1 and position 2. (Path C is a half circle.) Which path would result in the greatest displacement for a particle moving from position 1 to position 2?

• Path A
• Path B
• Path C
• All paths would result in the same displacement.

Stopwatch measures time t1

Stopwatch measures time t2

(average speed = total distance traveled / time to travel that distance)

Velocity

+x

• Velocity describes speed (magnitude) and direction
• Need time as well as position information
• Average velocity:

Park Ave. (reference point or origin)

William St. (point 1)

Central Ave. (point 2)

Stopwatch measures time t2

Stopwatch measures time t1

Velocity
• Average velocity could be positive or negative
• Trip from William St. to Central Ave. (previous picture) – assume t1= 0 s and t2= 40 s:
• Trip from Central Ave. to William St.:

+x

Park Ave. (origin)

William St. (point 2)

Central Ave. (point 1)

Trip #1

Trip #2

vaveis the same for both trips!

t1

t2

Velocity
• Assume t1 = 0 and t2 = 40 s:
• Ave. velocity depends only on the total displacement that occurs during time interval Dt= t2 – t1, not on what happens during Dt:

x

Central Ave.

William St.

time

Park Ave.

Velocity
• Note that each line represents how position changes with time and is not representative of a path in space
• Slope of line between 2 points on this graph gives average velocity between these points
• Instantaneous velocity = velocity at any specific instant in time or specific point along path
• Mathematically, it is the limit as Dt 0 of the average velocity:
• Say we wanted to know velocity at time t1
• Imagine moving t2 closer and closer to t1
• Dx & Dt become very small, but ratio not necessarily small

Slope = instant. velocity at timet2

Slope = instant. velocity at timet1

t1

t2

Instantaneous Velocity
• Graphically, instantaneous velocity at a point = slope of line tangent to curve at that point

x

Central Ave.

William St.

time

Park Ave.

Velocity
• Be careful not to interchange “speed” with “velocity”
• “speed” refers to a magnitude only
• “velocity” refers to magnitude and direction
• For example, I might run completely around a 1–mi. circular track in ¼ hour, with an average speed (round-trip) = dist. traveled / time = 1 mi / 0.25 hr = 4 mph
• BUT since
• In general: instantaneous speed = but average speed 

(I’m right back where I started!)

CQ2: The graph below represents a particle moving along a line. What is the total distance traveled by the particle from t = 0to t = 10seconds?

• 0 m
• 50 m
• 100 m
• 200 m

Slope = average acceleration

v2

v1

t1

t2

Acceleration
• Acceleration describes changes in velocity
• Rate of change of velocity with time
• Vector quantity (like velocity and displacement)
• “Velocity of the velocity”
• Average acceleration defined as:
• are the instantaneous velocities at t2 and t1, respectively
• To find average acceleration graphically:

v

t

Slope of tangent line = instantaneous acceleration

v1

t1

Acceleration
• We can also find the acceleration at a specific instant in time, similar to velocity
• Instantaneous acceleration:
• Graphically:
• BE CAREFUL with algebraic sign of acceleration – negative sign does not always mean that body is slowing down

v

t

Acceleration
• Example: Car has neg. acceleration if slowing down from pos. velocity, or if it’s speeding up in neg. direction (going in reverse)
• Must compare direction of velocity and acceleration:
• Be careful not to interchange deceleration with negative acceleration

Same sign

Oppositesign

(for example, going slower in reverse)

CQ3: Which animation shown in class gave the correct position vs. time graph? (Assume that the positive x direction is to the right.)
• Animation 1
• Animation 2
• Animation 3
• Animation 4

PHYSLET #7.1.1, Prentice Hall (2001)

Constant Acceleration

v

• For special case of constant acceleration, then v vs. t graph becomes a straight line
• Note that slope of above line (= acceleration) is the same for any line determined from 2 points on the line or for any tangent line since (for this special case)
• We can find the instantaneous acceleration from:

 Constant acceleration means constant rate of increase for v

t

Average value of v:

(since v is a linear function)

Also, by definition:

Setting (2) = (3), and using (1), we get:

Constant Acceleration
• Now let t1= 0 (can start our clock whenever we want) and t2 =t (some arbitrary time later), and define v(t1= 0)=v0 and v(t2= t) = v
• Graphical interpretation:

(takes form of equation of a line:y = b + mx)

v

v

at

v0

v0

t

Constant Acceleration
• We can use equations (1) and (4) together to obtain an equation independent of t and another equation independent of a using substitution, with the result:

(independent of t)

(independent of a)

Constant Acceleration
• Graphs of a, v, and x versus time:

Motion Graphs Interactive

CQ4: The graph below represents a particle moving along a line. When t = 0, the displacement of the particle is 0. All of the following statements are true about the particle EXCEPT:

• The particle has a total displacement of 100 m.
• The particle moves with constant acceleration from 0 to 5 s.
• The particle moves with constant velocity between 5 and 10s.
• The particle is moving backwards between 10 and 15 s.
Interactive Example Problem:Seat Belts Save Lives!

Animation and solution details given in class.

ActivPhysics Problem #1.8, Pearson/Addison Wesley (1995–2007)

Example Problem #2.39
• A car starts from rest and travels for 5.0 s with a uniform acceleration of +1.5 m/s2. The driver then applies the brakes, causing a uniform acceleration of –2.0 m/s2. If the brakes are applied for 3.0 s,
• how fast is the car going at the end of the braking period, and
• how far has the car gone?
• Solution (details given in class):
• 1.5 m/s
• 32 m
Example Problem #2.43
• A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 12 m/s, skates by with the puck. After 3.0 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 4.0 m/s2,
• how long does it take him to catch his opponent, and
• how far has he traveled in that time? (Assume that the player with the puck remains in motion at constant speed.)
• Solution (details given in class):
• 8.2 s
• 134 m
Free Fall
• Particular case of motion with constant acceleration: “free-fall” due to gravity
• Objects falling to the ground
• Galileo (16th Century) showed that bodies fall with constant downward acceleration, independent of mass (neglecting air resistance and buoyancy effects)
• Near the Earth’s surface, this acceleration has a constant value ofg 9.8 m/s2
• We can use all of the constant-acceleration equations for free fall
• Will use “y” instead of “x” when referring to vertical positions:
Free Fall

+y(up)

• Sign of depends on choice of +y – axis direction:
• Choice of axis not important – as long as you are consistent!
• always points toward the ground, however

 then =

0

0

 then =

+y(down)

CQ5: A ball is thrown straight up in the air. For which situation are both the instantaneous velocity and the acceleration zero?
• on the way up
• at the top of the flight path
• on the way down
• halfway up and halfway down
• none of these

+y

443 m

• y = y0 + v0t – ½ gt2
• – 443 m = 0 + 0 – ½ gt2
• 886 m / g = t2 t2 = 90.4 s2
• t =  9.5 s t = +9.5 s (neg. value has no physical meaning)
Free Fall Example Problem

How long would it take for Tom Petty to go “Free-Fallin’” from the top of the Sears Tower in Chicago? (Of course there is a safety net at the bottom.)

How fast will he be moving just before he hits the net?

• v = v0 – gt = 0 – gt = – (9.8 m/s2)(9.5 s) = – 93.1 m/s

(negative sign means downward direction)

CQ6: A pebble is dropped from rest from the top of a tall cliff and falls 4.9 m after 1.0 s has elapsed. How much farther does it drop in the next 2.0 seconds?

• 9.8 m
• 19.6 m
• 39 m
• 44 m
• 27 m

v

Free fall only

vtis called the “terminal velocity”

vt

• In reality, Tom Petty’s velocity would “tail off” and start reaching a maximum:
• Other notes about 1–D velocity and acceleration:
• In general, both velocity and acceleration are not constant in time
• Can use methods of calculus to generalize equations
• General formulas reduce to constant acceleration formulas when a(t)= constant =a
• Remember that formula d = vt only holds when v = constant!

Including air resistance effects

t