Transcript Chapter 2: Motion Along a Straight Line

Chapter 2: Motion Along a
Straight Line
AP Physics
Miss Wesley
Objectives
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In this chapter you will be able to have a
mathematical description of motion
For now, we don’t care what is causing the
motion.
For now, consider point-like objects (particle)
Definitions
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Position: the position of a particle can be specified
by some number along the x-axis.
–
Here x~3.7 m
-2
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-1
0
1
2
3
4
5
Displacement: The change in position of an object.
x = x2-x1
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Example: A particle moves from x1 = -2.0 m to x2 = 3.6 m.
Find the displacement.
x = x2-x1 = 3.6m – (-2m) = 5.6 m
Total Displacement = 5.6 m
Position vs. Time Graphs
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A convenient way to depict the motion of a particle.
Tells you the position of the particle at each instant in time.
Velocity:
vav = x/t
x = change in position (displacement)
t = change in time
x
Slope of this line = vav
t
Vav = (x2-x1)
(t2-t1)
Vav = slope of the line drawn between (t1, x1) and (t2, x2)
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Speed: the average speed is the total distance traveled by an object in a certain amount of
time.
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Note: sav ≠ vav
sAv = total distance
t
Velocity vs. Speed Example
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You drive down a road for 5.2 miles at 43
mph. You run out of gas and walk back to
the gas station 1.2 miles away in 30 minutes.
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A) What is vav for the trip?
B) What is the average speed for the trip?
The first step is to draw a picture:
Start
5.2 miles at 43 mph
1.2 miles
Finish at
Gas station
Velocity vs. Speed Example con’t
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A) Find vav:
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x = displacement from start to finish = 4.0 miles
Need to find t.
 tdriving
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= 5.2 miles/(43 mi/h) = 0.1209 h
twalking = 0.5 h
t = 0.1209 h + 0.5 h = 0.6209 h
Vav = 4.0 mi/0.6209 h = 6.44 mi/h
Velocity vs. Speed Example con’t
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B) Find average speed:
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Total distance = 5.2 miles + 1.2 miles = 6.4 miles
Total time = 0.6209 h (from A)
sav= 6.4 miles/0.6209 h
= 10.31 mi/h
Instantaneous Velocity
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How do we define the velocity of a particle at
a single instant?
When 2 points get close enough, the line
connecting them becomes a tangent line.
Instantaneous Velocity con’t
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Instantaneous Velocity: slope of the
tangent line to the x vs. t curve at a particular
instant.
v= instantaneous velocity = lim t0 = x/t = dx/dt
– This is the derivative of x with respect to t.
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Notations review:
v = instantaneous velocity
vav = v = average velocity
Acceleration
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When the instantaneous velocity is changing
with time, then it is accelerating
Average Acceleration:
aav = a = v/t = (v2-v1)/(t2-t1)
–
–
All of the velocities are instantaneous velocities.
This is the slope of the line connecting(t1, v1) and
(t2,v2) on a velocity vs. time graph
Instantaneous Acceleration
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Instantaneous acceleration is the slope of a
tangent line to a v vs. t curve at a particular
instant.
a = instantaneous acceleration = limt0 = v/t
Review
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Average Velocity:
vav = x/t = (x2-x1)
(t2-t1)
Average Speed:
sav = total distance/ t
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Instantaneous Velocity:
v = dx/dt = derivative of x with respect to t
– slope of the tangent line on an x vs. t graph
Review con’t
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Average acceleration:
Aav = v/t = (v2-v1)/(t2-t1)
– The velocities are the instantaneous velocities
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Instantaneous acceleration:
A = dv/dt = derivative of v with respect to t
OR – the slope of the tangent line on a v vs. t graph
OR  a = d2x/dt2 = the second derivative of x with
respect to t.
Brief Intro To Derivatives – Power Rule
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Supposex= ctn
Evaluate - dx / dt
dx/dt =nc*t(n-1)
Example 1: x=t2
dx/dt = 12t = v
Example 2: x= -5t3 +6t
dt/dt = v = -15t2+6
Brief Intro to Derivatives
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Example 3 – Suppose you know the height
of a ball as a function of time.
y(t)= -5(t-5) 2+125
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when t in seconds and y in meters.
a) Find the velocity as a function of time
b) Find the acceleration as a function of
time.
Example 3 – Derivatives – Chain Rule
a) Find the velocity as function of time
y(t)= -5(t-5) 2+125
y(t)= -5(t2 -10t+25)+125
y(t)= -5t2+50t
y’(t)= -10t+50= v(t)
OR
y(t)= -5(t-5) 2+125
y’(t)=-10(t-5) = -10t+50=v(t)
Example – Second Derivative
b) Find acceleration as a function of time.
Recall  v’(t)= a(t)
y’(t)= v(t) = -10t+50
v’(t)=-10=a(t)
Check Point - Acceleration
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Find sign (+, -) of acceleration if….
Speed Increasing
x
Speed Decreasing
Speed Increasing
x
x
Speed Decreasing
x
Motion with Constant Acceleration
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Acceleration does not change with time
SPECIAL CASE!
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It occurs often in nature (free fall)
Consider: a particle which moves along the
x-axis with constant acceleration a.
Suppose at time t = 0s its initial velocity is vo,
and its initial position is xo.
vo
xo
x
Motion with Constant Acceleration
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Using the previous
situation, find the velocity
at some time t.
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v
By definition:
a = Δv/Δt = (v-vo)/(t-0)
v = vo + at
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Graph will increase
linearly because only one
multiple of t.
Slope equals acceleration.
vo
t
Motion with Constant Acceleration
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Find the position at some later time.
Motion with Constant Acceleration
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Another handy equation:
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Square both sides of equation 1
Review – Special Case of Motion with
Constant Acceleration
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1.
2.
3.
Before you use these formulas, you MUST
make sure that the object has constant
acceleration
v = vo + at
x = xo + vot + ½at2
v2 = vo2 + 2a(x-xo)
The Acceleration of Gravity
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An example of motion with constant
acceleration
Experiments show that ALL objects fall to the
Earth with constant “free-fall” acceleration
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g = 9.81 m/s2
This means that heave objects fall at the
same rate as light objects (ignoring air
resistance)
Free Fall Motion
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We can use (1), (2), & (3) to describe free fall
motion with a few changes
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Because yes, it does have constant acceleration
y-axis is the direction of free-fall. It will point
upward.
a = -g because objects fall downward.
New Equations:
1.
v = vo – gt
2.
y = yo + vot - ½gt2
3.
v2 = vo2 - 2g(y-yo)
Example:
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A ball is released from rest from a height h.
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How long does it take to hit the ground?
DEMO!
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Choose a location in the room from which to
drop a ball.
Measure the height, and determine the
theoretical value for how long it should take
the ball to hit the ground
Measure how long it actually takes the ball to
hit the ground.
Calculate the percent error between the
measured time and the actual time
Example:
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A ball is released from rest from a height h.
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What is the ball’s velocity when it hits the ground?
(the instant before  when it actually hits the
ground the velocity will be zero)
Example 2
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A pitcher can throw a 100 mph fast ball. If he
throws the ball straight up, how long does it
take to reach the highest point?
Example 2 con’t
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What is the max. height?
Integration
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The inverse operation of taking the derivative
Recall:
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Given x(t), we can easily find v(t)
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V(t) is equal to dx/dt
Suppose we are given v(t), how can we find
the displacement (Δx) between ta and tb?
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USE INTEGRATION!
Velocity vs. Time Graph
Integration con’t
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We can make the equation exact by taking
the limit as Δti  0
Δx = lim Δti 0 Σi vi Δti = ∫tatb vdt
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“The itegral of vdt between ta and tb”
Another interpretation:
Δx = “area under the v vs. t curve”