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Lecture 3Lecture 3
Vectors and scalar quantities.Motion in Two Dimensions.Relative Motion.
What is a vector?What is a vector?
• A scalar quantity is one that is represented by a single number. (Eg: Mass, length, time, temperature, volume.)
• A vector is a quantity which has both magnitude and direction. (Eg: Displacement, velocity, force.)
Direction: something like “Makes an angle of 36 with the horizontal as measured CCW”
Notation: A or A (or A)
Magnitude (how long): |A| or A
• Equal vectors: Moving from P to Q, and from R to S:
x Qx Q
P x P x x Sx S
R x R x
AA
BBA = B
Vector basics
x Qx Q
P x P x
AA
-A-A
•Opposite vectors: Moving from P to Q, and from Q to P.
•Unit vector (Â): magnitude equals one. A= A Â
Vector addition: Moving from P, to Q, to R
x Qx Q
P x P x
x Rx R
AA
BB
CC
AA
BB
C =A-B
`
CC
CC-B-B
C =A+B
Vector subtraction: It’s an addition!
AB = A +( B)
Visually: What do I have to add to B to get A?
ACT: Vector additionACT: Vector addition
All the vectors below have the same magnitude. Which of the following arrangements will produce the largest resultant when the two vectors are added?
A. B. C.
ComponentsComponents
A vector can be thought of as the vector-sum of the projections along the coordinate axes.
x
y
A
Ax
Ay
A⃗ =A⃗x + A⃗y + (A⃗z)
Polar Components (only 2D)Polar Components (only 2D)
Magnitude |A| and direction θ
Ax=|A|cos θAy=|A|sin θ
|A|=√Ax2+Ay
2
θ=arctan(Ay
Ax)
θ
A|A|
angle θ taken counterclockwise from the x axis
Vector addition in terms of componentsVector addition in terms of components
• Geometric:
A
BC A+B=C
• Algebraic: Ax+Bx=Cx
Ay+By=Cy
Az+Bz=Cz
i.e., do all the walking in the x-direction first, then all the walking in the y-direction, etc.
Vector addition in terms of componentsVector addition in terms of components
• Algebraic: v1x+v2x=vx
v1y+v2y=vy
v = |v⃗ | = √vx2+vy
2
vy
vx
= tan θ θ = tan−1 ( vy
vx)
2D (and 3D) motion2D (and 3D) motion
Now we need vectors to indicate position, velocity and acceleration, but the definitions we use in 1D are pretty much the same.
Position:
x
y
r⃗f
r⃗0
trajectory
Displacement:Δ r⃗ =r⃗ (t+Δ t )−r⃗ (t )
( or Δ r⃗ =r⃗f−r⃗i )Δr
r⃗ (t )
VelocityVelocity
instantaneous velecityv⃗ is always tangent
to the tragectory.
xr⃗ (t=3 s)
trajectory
r⃗ (t=1 s )
average: v⃗ave=Δ r⃗
Δ t
v⃗ (t=1 s )
v⃗ (t=3 s )v⃗ave points from intialposition toward finalposition.
v⃗ave
AccelerationAcceleration
average: a⃗ave =Δ v⃗
Δ t=
v⃗f−v⃗0
Δ t
Magnitude : a =|a⃗ave|
ax =Δ vx
Δ t ay =
Δ vy
Δ t
v⃗0
v⃗f
Δ v⃗ a⃗
v⃗f
v⃗0t +Δt
t
ACT: AccelerationACT: Acceleration
Shown below are the trajectory of a moving object and the snapshots taken every second. Which of the following is true about the components of the acceleration?
y
x1s
2s3s
4s
A) ax = 0, ay > 0 B) ax > 0, ay > 0 C) ax < 0, ay = 0
Δv
Note: Both the speed and the direction of velocity are changing!
y
x 1s2s
3s4s
v(1)v(1)v(2)
v(3)
v(1)v(3) a
Same equations in 1D except they are duplicated for each component.
Constant accelerationConstant acceleration in 2D in 2D
Important note: x and y equations above are totally independent:
vx = v0x + ax t
x = x0 + v0x +12
ax t 2
vx =vx+v0x
2vx
2 = v0x2 + 2 ax Δx
vy = v0y + a y t
y = y0 + v0y +12
ay t 2
vy =vy+v0y
2vy
2 = v0y2 + 2 ay Δy
v 2 = v02 +2 (axΔ x + ay Δ y )
Since vx2 + vy
2 = v 2
We can combine the two equations:
vx2 = v0x
2 + 2 ax Δx+ vy
2 = v0y2 + 2 ay Δy } =
Relative MotionRelative Motion
Sometimes an object has two velocities at the same time. Let’s say a person is walking on a train at 2 m/s in the opposite direction of the train’s motion at 8 m/s.
How fast this person is going relative to someone on the ground?
8 m/s -2 m/s 6 m/s+
vperson to train+vtrain to ground = vperson to ground
ACT: Relative motion of two carsACT: Relative motion of two cars Two cars A and B move along parallel tracks. Shown are the snapshots of their motion at 1s intervals.
A. Points to the right.
B. Points to the left.
C. Is zero.
At t = 2 s, the velocity of the green car with respect to the red car:
1s 2s 3s 4s
1s 2s 3s 4s
red
green
1s 2s 3s 4s
1s 2s 3s 4s
red
green
vR,bg
vG,bg
vG,bg = vG,R + vR,bg
vG,R = vG,bg — vR,bg
+ =
1s 2s 3s 4s
1s 2s 3s 4s
red
green
Position of green relative to red:
1 2 3 4 t (s)
xG,R
overtaking
Negative slope
Negative relative velocity
A. Kid A
B. Kid B
C. Tie
D. Depends on the ratio vk/vsw
E. Depends on the sign of vsw
Example: Airport raceExample: Airport race
Two bored kids at the airport decide to race. Both kids walk with speed vk. One kid (A) will walk on the ground while the other (B) will walk on the “moving sidewalk” that moves with speed vsw. The race is roundtrip. Which kid wins the race?
Two bored kids stuck at the airport (flight delays) decide to race. Both kids walk with speed vk. One kid (A) will walk on the ground while the other (B) will walk on the “moving sidewalk” that moves with speed vsw. The race is roundtrip. Which kid wins the race?
Time for roundtrip, kid A:Let d = length of the moving sidewalk.tA =
2dvk
Time for roundtrip, kid B: B against SW with SWt t t= +
vkid B relative to ground = vk−vswvkid B relative to ground = vk+vsw
tB =d
vk−v sw
+d
vk+v sw
=2 vk dvk
2−vsw2
=2 dvk ( 1
1−vsw
2
vk2 )=tA ( 1
1−vsw
2
vk2 )
If vsw<vk , then 1
1−vsw
2
vk2
> 1, so tA<tB ( answer A)
vf,bg
2D relative motion
vf,bg
vt,bgvt,f
vt,f
vt,bgθ
From the film (a truck is moving across a film that is moving at half the speed of the truck): What is the angle between the path of the truck when the floor is not moving and its path when the floor is moving at half the speed of the truck?
vf,bg
2D relative motion
vf,bg
vt,bgvt,f
vt,f
vt,bgθ
From the film (a truck is moving across a film that is moving at half the speed of the truck): What is the angle between the path of the truck when the floor is not moving and its path when the floor is moving at half the speed of the truck?
So the trajectory makes an angle q = tan1 (0.5) = 26.6 clockwise from the vertical.
θv
v/2
2tan 0.5v
vq = =
EXAMPLE: BoatEXAMPLE: Boat
A boat which engine can make it move at 5.0 m/s relative to the water is trying to go across a 100-m wide river to a point on the opposite shore and right North of its starting position. The river flows due West at 3.0 m/s. How long does the trip take?
N
S
EW
EXAMPLE: BoatEXAMPLE: Boat
A boat which engine can make it move at 5.0 m/s relative to the water is trying to go across a 100-m wide river to a point on the opposite shore and right North of its starting position. The river flows due West at 3.0 m/s. How long does the trip take?
N
S
EWvb,g
vw,g
vb,w
vb,g = vb,w + vw,g
vb,g
3.0 m/s
5.0 m/s
vb,g = √(5.0 m/s)2− (3.0 m/s)2= 4.0 m/s
Δ t =Δ xg
vb,g
= 100 m4.0 m/s
=25 s
Faster than light?Faster than light?
Special relativity (postulated by A. Einstein): Nothing can travel faster that light in vacuum (c ~ 3 × 108 m/s) in any frame of reference.
The fix: Galilean transformations are OK for speeds v << c only. For high speeds, we need to use the so-called Lorentz transformations (which become the Galilean transformation in the limit of small v).
Two spaceships moving at 200,000 km/s relative to Earth head toward each other somewhere in the galaxy . What is their relative speed?
Two cars driving at 50 mph head toward each other on a highway. What is their relative speed? Answer: 100 mph
Answer: 400 000 km/s
This is faster than light (300 000 km/s) !! Is this possible??? No.