Physics I
95.141
LECTURE 3
9/13/10
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Exam Prep Question
• 2 cars are racing. Car A begins accelerating (aA=4m/s2), but
Car B stalls. After 5s, Car A thinks it has won and stops
accelerating, while Car B finally starts and accelerates at a
rate of aB=5m/s2.
(a) (5pts) What is the speed of Car A when Car B finally starts moving?
(b) (5pts) What is the head start (in m) that Car A gets?
(c) (10 pts) How long (in s) until Car B catches up to Car B?
(d) (10 pts) What is the minimum length of the race track required for
Car B to win the race?
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Exam Prep Question
2 cars are racing. Car A begins accelerating (aA=4m/s2), but
Car B stalls. After 5s, Car A thinks it has won and stops
accelerating, while Car B finally starts and accelerates at a
rate of aB=5m/s2.
– Draw Diagram/Coord. System
– Knowns and unknowns
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Exam Prep Question
2 cars are racing. Car A begins accelerating (aA=4m/s2), but
Car B stalls. After 5s, Car A thinks it has won and stops
accelerating, while Car B finally starts and accelerates at a
rate of aB=5m/s2.
(a) (5pts) What is the speed of Car A when Car B finally starts moving?
(b) (5pts) What is the head start (in m) that Car A gets?
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Exam Prep Question
2 cars are racing. Car A begins accelerating (aA=4m/s2), but
Car B stalls. After 5s, Car A thinks it has won and stops
accelerating, while Car B finally starts and accelerates at a
rate of aB=5m/s2.
(c) (10 pts) How long (in s) until Car B catches up to Car B?
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Exam Prep Question
2 cars are racing. Car A begins accelerating (aA=4m/s2), but Car
B stalls. After 5s, Car A thinks it has won and stops accelerating,
while Car B finally starts and accelerates at a rate of aB=5m/s2.
(d) (10 pts) What is the minimum length of the race track required for Car B
to win the race?
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Outline
•
•
•
•
•
Freely Falling Body Problems
Vectors and Scalars
Addition of vectors (Graphical)
Adding Vectors by Components
Unit Vectors
• What Do We Know?
–
–
–
–
Units/Measurement/Estimation
Displacement/Distance
Velocity (avg. & inst.), speed
Acceleration
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Review of Lecture 2
• Last Lecture (2) we discussed how to describe
the position and motion of an object
•
•
•
•
•
Reference Frames
Position
Velocity
Acceleration
Constant Acceleration
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
v  vo  at
1 2
x  xo  vot  at
2
v 2  vo2  2a ( x  xo )
v  vo
v
2
Freely falling Bodies
• Most common example of constant acceleration is a
freely falling body.
• The acceleration due to gravity at the Earth’s surface is
basically constant and the same for ALL OBJECTS
(Galileo Galilei)
m
a g  g  9.8
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
s2
Example Problem
• Batman launches his grappling bat-hook upwards, if the beam it
attaches to is 50m above Batman’s Batbelt, at what bat-velocity
must the hook be launched at in order to make it to the beam?
(Ignore the mass of the cord and air resisitance)
1) Choose coordinate system
2) Knowns and unknowns
3) Choose equation(s)
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Example Problem
• Batman launches his grappling bat-hook upwards, if the beam it
attaches to is 50m above Batman’s Batbelt, at what bat-velocity
must the hook be launched at in order to make it to the beam?
v  vo  at
(Ignore the mass of the cord and air resisitance)
1
3) Choose equation(s)
x  xo  vot  at 2
2
4) Solve
2
2
v  vo  2a ( x  xo )
v
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
v  vo
2
Vectors and Scalars
• A quantity that has both direction and
magnitude, is known as a vector.
– Velocity, acceleration, displacement, Force,
momentum

– In text, we represent vector quantities as v
 
,a ,r
• Quantities with no direction associated with them
are known as scalars
– Speed, temperature, mass, time
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Vectors and Scalars
• In the previous chapter we dealt with motion in a
straight line
– For horizontal motion (+/- x)
– For vertical motion (+/- y)
• Velocity, displacement, acceleration were still
vectors, but direction was indicated by the sign
(+/-).
• We will first understand how to work with vectors
graphically
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Vectors
• Graphically, we can depict a vector as an arrow
– Arrows have both length (magnitude) and direction.
5
y-axis
x-axis
-5
5
-5
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Addition of vectors
• In one dimension
– If the vectors are in the same direction
– But if the vectors are in the opposite direction
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Addition of Vectors (2D)
• In two dimensions, things are more complicated
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Addition of Vectors
• “Tip to tail” method
– Draw first vector
– Draw second vector, placing tail at tip of first vector
– Arrow from tail of 1st vector to tip of 2nd vector is
5

D1  3 m ( x )

D2  2 m ( y )
y-axis
x-axis
-5
5
-5
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Commutative property of vectors
• “Tip to tail” method works in either order
– Draw first vector
– Draw second vector, placing tail at tip of first vector
– Arrow from tail of 1st vector to tip of 2nd vector is
5

D1  3 m ( x )

D2  2 m ( y )




D1  D2  D2  D1
y-axis
x-axis
-5
5
-5
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Three or more vectors
• Can use “tip to tail” for more than 2 vectors
+
=
+
5
y-axis
x-axis
-5
5
-5
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Subtraction of vectors


• For a given vector V1 the negative of the vector  V1 is a
vector with the same magnitude in the opposite
direction.
 


V1  V2  V1  (V2 )
-
=
+
• Difference between two vectors is equal to the sum of
the first vector and the negative of the second vector
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Adding vectors by components
• Adding vectors graphically is useful
to understand the concept of
vectors, but it is inherently slow (not
to mention next to impossible in
3D!!)
• Any 2D vector can be decomposed
into components
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Determining vector components
• So in 2D, we can always write any vector as the sum of
a vector in the x-direction, and one in the y-direction.
• Given V(V,θ), we can find Vx and Vy
 

V  Vx  V y
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Determining vector components
V  Vx  V y
2
• Or, given Vx and Vy, we can find V(V,θ).
  tan
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Vy
Vx
2
Example
• A vector is given by its vector components:


Vx  2 , V y  4
• Write the vector in terms of magnitude and
direction
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Adding vectors by components
• Given V1 and V2, how can we find V= V1 + V2?
  
V  V1  V2
 V1 x  V2 x   V1 y  V2 y 
 Vx  V y
V
V1
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
V2
3D Vectors
• Adding vectors vectors by components is especially
helpful for 3D vectors.


V1  V1 x  V1 y  V1z , V2  V2 x  V2 y  V2 z
  
V  V1  V2  V1 x  V2 x   V1 y  V2 y   V1z  V2 z 
• Also, much easier for subtraction


V1  V1 x  V1 y  V1z , V2  V2 x  V2 y  V2 z
  
V  V1  V2  V1 x  V2 x   V1 y  V2 y   V1z  V2 z 
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Multiplying a vector by a scalar
• You can also multiply a vector by a scalar

cV1
• When you do this, you don’t change the direction of the
vector, only its magnitude
c=2
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
c=4
c=-2
Unit Vectors
• Up to this point, we have written vectors in terms of their
components as follows:
• There is an easier way to do this, and this is how
we will write vectors for the remainder of the
course:

V  V x iˆ  V y ˆj  Vz kˆ
iˆ, ˆj , kˆ known as unit vectors
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Unit Vectors
• What are unit vectors?
– Unit vectors have a magnitude of 1 and point along major axes
of our coordinate system
• Writing a vector with unit vectors
is equivalent to multiplying each
unit vector by a scalar



Vx  Vx iˆ , V y  V y ˆj , Vz  Vz kˆ

V  Vx iˆ  V y ˆj  Vz kˆ
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Unit Vectors
• For a vector with components:



Vx  4 , V y  3 , Vz  2
• Write this in unit vector notation
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Example: Vector Addition/Subtraction
•
Displacement
– A hiker traces her movement along a trail. The first leg of her hike brings her to
the foot of the mountain:
1
– On the second leg, she ascends the mountain, which she figures to be a

displacement of:
ˆ
ˆ
ˆ

V  2500m iˆ  500mˆj
V2  500mi  700mj  700mk
–
–

On the third, she walks along a plateau. V  600mˆ
j
3

Then she falls of a cliff
V4  500m kˆ
– What is the hiker’s final displacement?
95.141, F2010, Lecture 3
Department of Physics and Applied Physics
Example: Vector Addition/Subtraction

V1  2500m iˆ  500mˆj

V3  600mˆj

V2  500miˆ  700mˆj  700mkˆ

V4  500m kˆ
95.141, F2010, Lecture 3
Department of Physics and Applied Physics