Transcript Two-Dimensional Motion and Vectors

Two-Dimensional Motion and
Vectors
Honors Physics
Unit 2
• Our last unit focused on motion in a straight
line (usually forward and backward) using +
and – signs to indicate direction
• This unit focuses on motion that is not in a
straight line
(Re)Introduction to Vectors
• Vectors indicate direction, Scalars do not
**Symbol Change Alert**
• Sometimes we use the same symbol to represent
2 different things. (velocity and speed = v)
Velocity is a vector quantity and Speed is scalar
• Your book will put vector quantities in bold (v)
and scalar quantities in italics (v)
• You can distinguish between the two when you
write them with an arrow above the symbol.( )
Vectors
• Remember: In diagrams, vectors are shown as
arrows that point in the direction of the
vector.
– The length of the vector arrow is proportional to
it’s magnitude
Resultants
• The sum of two or more vectors is called the
Resultant.
– make sure that you have the same units when
adding vectors and that you are describing similar
quantities (ex. You wouldn’t add velocity and
displacement together, even though they are both
vector quantities. You wouldn’t add feet and
meters together either)
Adding Vectors Graphically
• If you walked 1600m (a) to a friends house,
then 1600m (b) to school, what is your total
displacement
You must draw the graph to
scale. For example, use 1cm
equals 50m
Using a ruler, you can
measure the length of the
resultant, and multiply the
length by the scale used to
get it’s magnitude.
Adding Vector Quantities
• Using a protractor in the last example can give
you a direction of the resultant.
• Measure the angle between the resultant and
the first vector
Remember Me?
• Head to tail method:
• Head to tail method: (More than 2 forces)
R
Head to Tail
• Vectors can be moved (added) in any order
Homework:
Problems 1-5
Page 85; Holt Physics
Determining Resultant Magnitude
and Direction
• For homework, we found magnitude and
direction using graphs
– Time consuming
– Accuracy
• There is a simpler way..
Magnitude of a resultant
• Pythagorean theorem
For any right triangle, the square of the hypotenuse
equals the sum of the squares of the other two sides,
or legs c2 = a2 + b2
y
d
Δy
Δx
x
Direction of a Resultant
• Tangent Function
For a right triangle the tangent of an angle is the ratio of
the opposite and adjacent sides(legs)
• It is knowing the angle that is important to
determining direction, so we can use the inverse
of the tangent function
Example Problem
• You climb on of Egypt’s pyramids that has a
height of 136m and a width of 2.30E2m. What
is the magnitude and direction of the
displacement for you as you climb from the
bottom to the top?
Given:
Unknowns:
Δy=136m
d=?
Δx = 115m
θ=?
• Use the Pythagorean theorem to find the
magnitude of your displacement
d2 = 1152 + 1362 d = 178m
• Now for direction use inverse tangent
θ=49.8°
More Practice
• A girl delivering newspapers travels three
blocks west, four blocks north and then six
blocks east.
1. What is her resultant displacement?
2. What is the total distance she travels
More Practice
• A quarterback takes the ball from the line of
scrimmage, runs backward for 10.0 yards, and
then runs sideways for 15.0 yards. At this
point, he throws a 50.0 yard pass straight
down the field. What is the magnitude of the
football’s resultant displacement?
More Practice
• A shopper pushes a cart 40.0m south down
one aisle and then turns 90.0° and moves
15.0m. He then makes another 90.0° turn and
moves 20.0m. Find the shopper’s total
displacement. (there can be more than one
correct answer)
Resolving Vectors into Components
• What do we mean by “components”?
– Horizontal part (x-axis) and Vertical part (y-axis)
y
x
If given the angle and the magnitude of
the vector force, its components can be
found using trigonometry
Vector Components Example
• Find the components of the velocity of a
helicopter traveling 95km/h at an angle of 35°
y
to the ground.
= 95km/h
Given:
= 95km/h
θ = 35°
Unknowns: Vx ; Vy
Step 1: Draw Diagram
Step 2: Use Trig to solve
Step 3: Check your work
Sin 35 = Vy/
Vy
35°
x
Cos 35 = Vx/
Vx
Vy = 54 km/h
Vx = 78 km/h
782 + 542 = 952 ??
Practice Problems
• How fast must a truck travel to stay beneath
an airplane that is moving 105km/h at an
angle of 25° to the ground?
• What is the magnitude of the vertical
component of the velocity of the plane?
Two More
• A truck drives up a hill with a 15° incline. If
the truck has a constant speed of 22m/s, what
are the horizontal and vertical components of
the truck’s velocity?
• What are the horizontal and vertical
components of a cat’s displacement when the
cat has climbed 5m directly up a tree?
Non Perpendicular Vectors
Homework
• Problem Workbook: Pg 18 (1-7) & Pg 22 (1-5)
Projectiles
• A projectile is an object that moves through
air and space with only gravitation force acting
on it.
• Projectiles have a horizontal and a vertical
component
• Horizontal component remains constant as
long as no other horizontal force is acting on it
Projectile
• The Vertical component has the same
properties of an object in free fall. (a constant
acceleration due to gravity)
• The two components (horizontal and vertical)
are independent of each other.
– Gravity has no effect on the horizontal velocity
and the horizontal velocity has no effect on the
acceleration due to gravity
Projectiles
• Projectiles launched at any Angle will fall
vertically from the ideal straight line, equally
– Launched Horizontally (Angle= 0°)
– Launched Above the Horizontal
– Launched Below the Horizontal
Projectiles
• Horizontal Projectile
5m
20 m
45 m
Projectiles
• Launched Above the Horizontal
5m
20 m
45 m
Projectiles
• Where did the 5m, 20m and 45m distances
come from?
– Gravity acting on the object
• Remember the vertical distance that an object falls
beneath the ideal straight line is the same as the
distance it would travel in a free fall
So, we can use the distance equation we had for objects
in a freefall
Δy = ½(g)t2 Since ‘g’ is constant (10.0m/s2) the
equation can be simplified to Δy = 5t2
Projectiles
• The path that projectiles follow is called a
parabola. The angle of the launch determines
the shape of the parabola
Projectiles
• You can find the Vertical and Horizontal
components of the object at any time (t) which
will let you solve for a resultant vector velocity.
• Ex) Stude tosses a ball into the air with a
horizontal component of 5m/s and a vertical
component of 30m/s. In 1 second intervals, find
the velocity components as the ball travels up
and the resultant velocities as the ball travels
downward
So, what about that frog…
• Givens: Δt = 0.60s v = 5.0m/s Δx = 2.4m
• Unknown: θ
1. Find the horizontal velocity
needed to leap 2.4m in 0.60s.
V = Δx/Δt =
4.0m/s
5.0m/s
2.4m
2. Using the horizontal
velocity and the
resultant velocity
calculate the angle
needed
5.0m/s
Cos-1 (4.0/5.0) =
36.8°
4.0m/s