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Chapter 2: One-Dimensional Motion. Motion at fixed velocity Definition of average velocity Motion with fixed acceleration Graphical representations. Displacement vs. position. Position: x (relative to origin) Displacement: D x = x f -x i. Average velocity. Average velocity. - PowerPoint PPT Presentation
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Chapter 2: One-Dimensional Motion
•Motion at fixed velocity•Definition of average velocity•Motion with fixed acceleration•Graphical representations
Displacement vs. position
Position: x (relative to origin)Displacement: x = xf-xi
basic formula
v =xt
=xf −xi
t
Average velocity
Average velocity
•Can be positive or negative•Depends only on initial/final positions•e.g., if you return to original position, average velocity is zero
basic formula
v =xt
=xf −xit
Instantaneous velocity
Let time interval approach zero
•Defined for every instance in time•Equals average velocity if v = constant•SPEED is absolute value of velocity
Graphical Representation of Average Velocity
Between A and D , v is slope of blue line
Graphical Representation of Instantaneous Velocity
v(t=3.0) is slope of tangent (green line)
Example 2.1
Carol starts at a position x(t=0) = 1.5 m.At t=2.0 s, Carol’s position is x(t=2 s)=4.5 mAt t=4.0 s, Carol’s position is x(t=4 s)=-2.5 m
a) What is Carol’s average velocity between t=0 and t=2 s?b) What is Carol’s average velocity between t=2 and t=4 s?c) What is Carol’s average velocity between t=0 and t=4 s?
a) 1.5 m/sb) -3.5 m/sc) -1.0 m/s
Example 2.2
On a mission to rid Spartan Stadium of vermin, an archer shoots an arrow across the stadium at an unlucky rat 200 meters away. The archer hears the squeal 2.2 seconds later. What was the velocity of the arrow? The speed of sound is 330 m/s.
Example 2.2: Visualize the problem!
Example 2.2
On a mission to rid Spartan Stadium of vermin, an archer shoots an arrow across the stadium at an unlucky rat 200 meters away. The archer hears the squeal 2.2 seconds later. What was the velocity of the arrow? The speed of sound is 330 m/s.
V = 125 m/s
Example 2.3a
The instantaneous velocityis zero at ___
A) aB) b & dC) c & e
Example 2.3b
The instantaneous velocity is negative at _____
A) aB) bC) cD) dE) e
Example 2.3c
The average velocity is zero in the interval _____ A) a-c
B) b-dC) c-dD) c-eE) d-e
Example 2.3d
The average velocity is negative in the interval(s)_________
A) a-b B) a-cC) c-eD) d-e
SPEED
• Speed is |v| and is always positive• Average speed is sum over |x| elements divided by elapsed time
Example 2.4
2
4
6
8
2 4 6 8 10 1200A
B
C
D
E
a) What is the average velocity between B and E?
b) What is the average speed between B and E?
a) 0.2 m/sb) 1.2 m/s
t (s)
x (m)
Acceleration
The rate of change of the velocity
Average acceleration: measured over finite time interval
Instantaneous acceleration: measured over infinitesimal interval, t -
> 0
a =vf −vit
Accelerometer Demo
Graphical Description of Acceleration
Acceleration is slope oftangent line in v vs. t graph
Graphical Description of Acceleration
a > 0
a < 0
a is positive/negativewhen v vs. t is
rising/fallingor when x vs t curves
upwards/downwards
Example 2.5a
a
b
cd
e
At which point(s) does the position equal zero?A) a onlyB) a and dC) b onlyD) b & d
Example 2.5b
a
b
cd
e
At which point(s) does the velocity equal zero?A) aB) b onlyC) c onlyD) b & dE) a & d
Example 2.5c
a
b
cd
e
At which point is the velocity negative?A) aB) bC) cD) dE) e
Example 2.5d
a
b
cd
e
At which segment(s) is the acceleration negative?
A) a-cB) c-dC) c-eD) d-e
Example 2.5e
a
b
cd
e
At which point(s) does the acceleration equal zero?
A) none of the belowB) bC) cD) dE) e
Constant Acceleration
• a vs. t is a constant• v vs t is a straight line• x vs t is a parabola
Eq.s of Motion
v f =v0 +at
x=12(v0 +vf )t
Solving Problems with Eq.s of Motion
5 variables: x, t, v0, vf, a
2 equations: v f =v0 +at
x=12(v0 +vf )t
3 variables must be given so that2 equations can solve for 2 unknowns
Example 2.6
Crash Houlihan speeds down the intersate, when she slams on the brakes and slides into a concrete barrier. The police measure skid marks to be 60 m long, and from a tape recording, know that she was breaking from 3.5 seconds. Furthermore, they know that her Mercedes would decelerate at 5.5 m/s2 while skidding. What was Crash’s speed when she hit the barrier?
7.52 m/s
Other Forms of Eq.s of Motion
x =v0 + (v0 + at)
2t
x = v0t +1
2at2
Substitute to eliminate vf
v f =v0 +at
x=12(v0 +vf )t
Other Forms of Eq.s of Motion
x =(v f − at) + v f
2t
x = v f t −1
2at2
Substitute to eliminate v0
v f =v0 +at
x=12(v0 +vf )t
Other Forms of Eq.s of Motion
x =(v0 + v f )
2
(v f − v0 )
a
ax=vf2
2−v02
2
Substitute to eliminate t
v f =v0 +at
x=12(v0 +vf )t
Final List of 1-d Equations
Which one should I use?Each Eq. has 4 of the 5 variables: x, t, v0, v & aAsk yourself “Which variable am I not given and not interested in?” If that variable is t, use Eq. (5).
basic equations:
1) v =v0 +at
2) x=12(v0 +v)t
3) x=v0t+12at2
4) x=vft−12at2
5) ax=vf2
2−v02
2
Example 2.6 (Revisited)
Crash Houlihan speeds down the intersate, when she slams on the brakes and slides into a concrete barrier. The police measure skid marks to be 60 m long, and from a tape recording, know that she was breaking from 3.5 seconds. Furthermore, they know that her Mercedes would decelerate at 5.5 m/s2 while skidding. What was Crash’s speed when she hit the barrier?
7.52 m/s
Example 2.7aA drag racer starts her car from rest and accelerates at 10.0 m/s2 for the entire distance of a 400 m (1/4 mi) race.How much time was required to finish the race?
a) v =v0 +at
b) x=12(v0 +v)t
c) x=v0t+12at2
d) x=vft−12at2
e) ax=vf2
2−v02
2
Example 2.7bA drag racer starts her car from rest and accelerates at 10.0 m/s2 for the entire distance of a 400 m (1/4 mi) race.What was her final speed?
a) v =v0 +at
b) x=12(v0 +v)t
c) x=v0t+12at2
d) x=vft−12at2
e) ax=vf2
2−v02
2
Example 2.7cA drag racer starts her car from rest and finishes a race in 3.5 seconds with a constant acceleration for the entire distance of a 400 m (1/4 mi) race.What was her final speed?
a) v =v0 +at
b) x=12(v0 +v)t
c) x=v0t+12at2
d) x=vft−12at2
e) ax=vf2
2−v02
2
Free Fall
• Objects under the influence of gravity (no resistance) fall with constant downward acceleration (if near Earth’s surface).
g = 9.81 m/s2
• Use the usual equations with a --> -g
•Father was a musician, experimented with music•Initially was a professor teaching pre-meds•Developed telescope ~ 1610:
Milky Way = starsMoons of JupiterPhases of Venus…
•Measured g•Quantified mechanics•In 1632, published Dialogue concerning the two greatest world systems•Was found guilty of heresy
Galileo
Example 2.8a
A man drops a brick off the top of a 50-m building. The brick has zero initial velocity.
a) How much time is required for the brick to hit the ground?
b) What is the velocity of thebrick when it hits the ground?
A B
c
a) 3.19 sb) -31.3 m/s
Example 2.8b
A man throws a brick upward from the top of a 50 m building. The brick has an initial upward velocity of 20 m/s.
a) How high above the building does the brick get before it falls?
b) How much time does the brick spend going upwards?
c) What is the velocity of the brick when it passes the man going downwards?
d) What is the velocity of the brick when it hits the ground?
e) At what time does the brick hit the ground?
A B
c
Example 2.8b
A man throws a brick upward from the top of a 50 m building. The brick has an initial upward velocity of 20 m/s.
a) How high above the building does the brick get before it falls?
b) How much time does the brick spend going upwards?
c) What is the velocity of the brick when it passes the man going downwards?
d) What is the velocity of the brick when it hits the ground?
e) At what time does the brick hit the ground?
a) 20.4 mb) 2.04 sc) -20 m/sd) -37.2 m/se) 5.83 s
Example 2.9a
A man throws a brick upward from the top of a building. (Assume the coordinate system is defined with positve defined as upward)
At ‘A’ the acceleration is positive
A C
D
A C
D
B
E
a) Trueb) False
Example 2.9b
A man throws a brick upward from the top of a building. (Assume the coordinate system is defined with positve defined as upward)
At ‘B’ the velocity is zero
A C
D
A C
D
B
E
a) Trueb) False
Example 2.9c
A man throws a brick upward from the top of a building. (Assume the coordinate system is defined with positve defined as upward)
At ‘B’ the acceleration is zero
A C
D
A C
D
B
E
a) Trueb) False
Example 2.9d
A man throws a brick upward from the top of a building. (Assume the coordinate system is defined with positve defined as upward)
At ‘C’ the velocity is negative
A C
D
A C
D
B
E
a) Trueb) False
Example 2.9e
A man throws a brick upward from the top of a building. (Assume the coordinate system is defined with positve defined as upward)
At ‘C’ the acceleration is negative
A C
D
A C
D
B
E
a) Trueb) False
Example 2.9f
A man throws a brick upward from the top of a building. (Assume the coordinate system is defined with positve defined as upward)
The speed at ‘C’ and at ‘A’ are equal
A C
D
A C
D
B
E
a) Trueb) False
Example 2.9g
A man throws a brick upward from the top of a building. TRUE OR FALSE. (Assume the coordinate system is defined with positve defined as upward)
The velocity at ‘C’ and at ‘A’ are equal
A C
D
A C
D
B
E
a) Trueb) False
Example 2.9h
A man throws a brick upward from the top of a building. (Assume the coordinate system is defined with positve defined as upward)
The speed is greatest at ‘E’
A C
D
A C
D
B
E
a) Trueb) False
