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Vectors

Vectors

A vector has magnitude and direction (eg. velocity)

A scalar only has magnitude (eg. speed)

Examples of vectors are displacement, velocity, momentum force and field.

Vectors in One-Dimension

In one dimension we use the convention of up as positive and down as negative; right as positive and left as negative.

Vector Addition

For example, we can add force vectors to determine the resultant force on an object:

The tip-to-tail method of adding vectors (arrows) best illustrates vector addition:

We could also add separate displacement vectors of an object to determine its final displacement:

5 N

5 N 3 N 2 N

(+5) N (−3) N (+2) N

2 N (−3) N

Displacement Vector

4

3v⃑1

v⃑2∑ v⃑4

3

∑ v⃑

√72+72

: angle measured from the origin of the vector

Adding Vectors in 2-Dimensions

Σ sigma=∑ of

Example 1: Calculate the magnitude of ∑ v⃑ (resultant vector).

x y |v⃑|v⃑1 4 3 5v⃑2 3 4 5Σ 7 7 7√2

OR

∑ v⃑x= v⃑1x+ v⃑2x=4+3=7 ¿ ∑ v⃑ y= v⃑1 y+ v⃑2 y=3+4=7

∴ ∑ v⃑=√72+72=√2 (72)=7√2≐ 9.90

Example 2: Resolving vectors into x & y components. Calculate |∑ v⃑|.

Resolve v⃑1 into v⃑1x and v⃑1 y:

v⃑1y

5=sin 30 ° (sinθ= O

H)

v⃑1 y=5 sin30 °=5× 0.5=2.5

v⃑1x

5=cos 30° (cos θ= A

H)

v⃑1x=5 cos30 °=5 × √32

=5√32

≐ 4.33

Repeat this process for v⃑2:

4

3

4 3

v⃑2 y=7 sin 60 °=7× √32

=6.06

Calculate the total horizontal and vertical displacements to find ∑ v⃑:

∑ v⃑x=4.33+3.5=7.83 ∑ v⃑ y=2.5+6.06=8.56

∑ v⃑=√(∑ v⃑ x)2+(∑ v⃑ y)

2=√ (7.83 )2+(8.56)2=11.6

Finding θ:

tanθ=∑ v⃑ y

∑ v⃑x

=8.567.83

∴ θ=tan−1( 8.567.83 )=47.7 °

Practice on Components

1. Find ∑ d⃑ and θ:

∑ d⃑=√ (d⃑ x)2+ (d⃑ y )2

¿√ (15 m )2+(6 m)2=16.2 m

tanθ= 6 m15 m

=2.5

∴ tan−1(2.5)=21.8 °

2. Find p⃑x and p⃑y:

p⃑x=(17 kg ∙ m /s ) (cos 50° )=10.9 kg ∙ m /s

p⃑y= (17 kg ∙m /s ) (sin 50 ° )=13.0 kg ∙ m /s

3. Find ∑ F⃑and F⃑ x:

sin 60 °= 25 N

∑ F⃑∴ ∑ F⃑= 25 N

sin 60 °=28.9 N

tan 60°=25 NF⃑x

∴ F⃑x=25 N

tan 60°=14.4 N

Example 2 - Another Way: Sine/Cosine Laws Method

Sine Law

b

c a

CA

B

sin Aa

= sin Bb

= sin Cc

Cosine Law

a2=b2+c2−2 bc cos A

a2=b2+c2−2 bc cos A ∴ (∑ v⃑ )2=52+72−2 (5 )(7)cos150 ° ∴ ∑ v⃑=11.6

sin Aa

= sinCc

∴ sin 150°11.6

= sin C7

∴ ∠C=sin−1( 7 sin 150 °11.6 )=17.6 °

θ=17.6 °+30°=47.6 °

Vector Subtraction

An example of vector subtrction is “relative velocity.”

How fast does Margaret seem to be moving relative to Bob?

v⃑MB=v⃑ M−v⃑ B ⇒ object ' svelocity−observer ' s velocity

¿ (+30 ) km /h−(+50 ) km /h= (−20 ) km /h ⇒ ¿be moving towards Bob .¿

We will see that writing it in this form makes it easier when solving 2-D problems:

v⃑MB

←= v⃑ M

→+(− v⃑B)⟵

Using the tip-to-tail method for vector addition,

A=150°

a=∑ v⃑=?

b=5c=7

Add the following vectors graphically:

Example 1: Sergey runs off the edge of the pool and dives into it. He runs with a speed of 6.0 m/s and descends a displacement of −2.0 m.

a) Calculate his time of flight after leaving the ground.

¿ y>¿ d⃑ y=v⃑ i y t+ 12

a⃑ y t 2=12

a⃑ y t 2(v⃑ i y=0 m /s)

t=√ 2 d⃑ y

a⃑y

=√ 2 (−2.0 m )−9.8 m /s2=0.64 s

b) Calculate his horizontal displacement from the edge of the pool to the water surface.

¿ x>¿ d⃑ x= v⃑ x t=(6.0 m /s ) (0.64 s )=3.8 m ( v⃑x is constant for all projectile motion)

c) Calculate his velocity as he enters the water. (Note: this is NOT 0 m/s)

¿ y>¿ v⃑ y=v⃑ i y+ a⃑y t= a⃑y t=(−9.8 m /s2 ) (0.64 s)=−6.3 m /s

¿ x>¿ v⃑ f x= v⃑ x=6.0 m /s

∑ v⃑=√( v⃑x )2+( v⃑ y )2=√ (6.0 m /s )2+ (−6.3 m /s )2=8.7 m /s

tanθ=v⃑ y

v⃑x

=−6.3 m /s6.0 m /s

θ=tan−1(−6.3 m/ s6.0 m/ s )=−46°

v⃑ f=8.7 m /s , 46 °below the horizontal

The Most Challenging Par-3 in the World

The golf ball is hit off the tee at 60.0 m/s at an angle of incline of 10.0° to the horizontal. Given that the tee-off is elevated 430 m above the flag and a horizontal distance of 650 m from it, calculate

a) How far from the flag the ball lands.

Method: Find the time of flight of the golf ball, and use that to calculate its horizontal displacement. Finally, determine the difference between the positions of the golf ball and the flag.

¿ y>¿ d⃑=v⃑ i t+12

a⃑ t 2 ⇒ −430m=(60.0m /s ) (sin10 ° ) t+ 12

(−9.80 m /s2) t 2

( 4.90m /s2) t 2−(60.0m /s ) (sin 10.0° ) t−430m=0

t=−[−(60.0 m /s ) (sin 10 ° ) ] ±√[−(60.0 m / s) (sin 10.0 ° ) ]2−4 ( 4.90m /s2 ) (−430 m )

2 ( 4.90m /s2 )

t=10.5 s ¿ t=−8.36 s (reject)

¿ x>¿ d⃑ x=v⃑ x t=(60.0m /s ) (cos10.0 ° ) (10.5 s)=620 m

Distance ¿ Flag=650 m−620 m=30 m

b) The velocity of the ball just prior to hitting the ground.

Method: Find v⃑xf and v⃑ yf . The final velocity is the resultant of these two vectors.

¿ x>¿ v⃑ f x= v⃑ ix= v⃑ cosθ=(60.0 m /s ) (cos 10.0° )=59.1m / s

¿ y>¿ v⃑ f2= v⃑ i

2+2 a⃑ d⃑ ⇒ v⃑ f y=−√ v⃑ i y2+2a⃑ y d⃑ y

v⃑ f y=−√[ (60.0 m / s ) (sin 10.0 ° ) ]2+2 (−9.80 m /s2 ) (−430 m )=−92.4 m/ s

Final Velocity=∑ v⃑=√( v⃑x )2+( v⃑ y )2=√ (59.1m /s )2+(−92.4 m /s )2=110m / s

tanθ=v⃑ y

v⃑x

=−92.4 m / s59.1 m /s

∴ θ=tan−1(−92.4 m / s59.1 m /s )=−57.4 °(57.4 ° belowthe horizontal)

c) Determine the maximum height above the ground reached by the ball.

At maximum height, v⃑ y=0m /s

since the ball is moving

horizontally.

¿ y>¿ v⃑ f2= v⃑ i

2+2 a⃑ d⃑

0m /s= [(60.0m / s) (sin 10.0 ° ) ]2+2 (−9.80m /s2) d⃑ y

d⃑ y=−[ (60.0 m / s) (sin 10.0 ° ) ]2

2 (−9.80 m /s2 )=5.54 m

Total Height=430 m+5.54m=435.54 m≐ 436 m

The “Range” Equation

“Range” in projectile motion is the horizontal displacement of a projectile across a level surface.

Condition : d⃑ y=0 m

Range , d⃑x=− v⃑ο

2sin 2 θg⃑

where g⃑=−9.80 m /s2

Hint :Trigonometric Identity sin 2 θ=2 sin θ cosθ

Resolving v⃑ο into v⃑οx and v⃑ο y:

v⃑ο y= v⃑οsin θ ¿ v⃑ο x= v⃑οcos θ (constant horizontal velocity)

¿ y>¿ d⃑ y=v⃑ο y t+ 12

a⃑y t2 ⇒ 0= v⃑ο sinθ t + 12

g⃑ t2

t (v⃑ οsin θ+ 12

g⃑ t)=0 ∴ t=0 ¿ t=−2 v⃑ο sin θ

g⃑

¿ x>¿ d⃑ x=v⃑ο x t= v⃑ο cosθ t ⇒ Substitute∈t=−2 v⃑ οsin θ

g⃑

d⃑ x=v⃑ οcosθ (−2 v⃑ο sin θg⃑ )=− v⃑ο

2 (2 cosθ sin θ )g⃑

∴ Range ,d⃑ x=−v⃑ο

2 sin 2θg⃑

Example: A fireman points his hose at a blazing forest fire. He keeps a distance of 30 m from the base of the fire. Given that the water exits the hose with a speed of 40 m/s, what angle should he aim the hose?

d⃑ x=− v⃑ο

2 sin 2θg⃑

⇒ sin 2θ=− d⃑x g⃑

v⃑ο2 =

−(30 m ) (−9.80m /s2 )( 40 m/ s )2

=0.18375

2 θ=sin−1 (0.18375 )=10.588 °∧169.412 ° ∴ θ=5.3 °∧84.7 °

Chapter 3, Problem 71: Agent Tim, flying a constant 185 km/h horizontally in a low-flying helicopter, wants to drop a small explosive onto a master criminal’s automobile travelling 145 km/h on a level highway 88.0 m below. At what angle (with the horizontal) should the car be in his sights when the bomb is released?

Projectile’s Motion

¿ y>¿ d⃑ y=v⃑ο y t+ 12

a⃑y t2=12

a⃑y t2(∵ v⃑οy=0)

−88.0 m=12

(−9.80 m /s2) t 2

t=4.24 s

¿ x>¿ d⃑ x= v⃑ο x t

d⃑ x=( 185 kmh

×1h

3600 s×

1000 m1 km ) (4.24 s )

d⃑ x=217.8 m

Car’s Motion

¿ x>¿ D⃑x=V⃑ x t=( 145 kmh

×1 h

3600 s×

1000 m1 km ) (4.24 s)=170.7 m

Initial Difference in Horizontal Distance:

∆ d⃑=d⃑ x−D⃑x

∆ d⃑=217.8 m−170.7 m=47.1 m

Initial Angle of Sighting:

tanθ=|d⃑ y||∆ d⃑|

=88.0 m47.1m

θ=61.8 °below the horizontal

Actual, Resultant, and Relative Velocities

Direct Notation

Note:

Northeast implies 45° N of E. Northwest implies 45° N of W.

Southeast implies 45° S of E. Southwest implies 45° S of W.

Resultant (Actual) Velocity

Example: An airplane travels with an airspeed of 600 km/h and a heading due west. A wind current moves with a speed of 70 km/h 30° S of W.

Airspeed: speed of plane in “still air,” when there’s no wind

Heading: direction of travel in still air

a) Calculate the resultant velocity of the airplane.

Note : ¿ ground observer¿ the wind (ie . velocity of plane∈still air )¿ (ie . resultant∨actual velocity of the plane)

¿¿

RelativeVelocity : v⃑ AB⏟Object Relative¿

Observer= v⃑ A⏟Object

− v⃑B⏟Observer

¿

v⃑PW= v⃑PG−v⃑WG ⇒ v⃑ PG= v⃑PW + v⃑WG

where v⃑PG is the resultant velocity of the plane.

v⃑WGx=−70 cos 30°=−60.62 km /h

v⃑WGy=−70 sin 30 °=−35 km /h

v⃑PG x=v⃑ PW x+v⃑WGx

v⃑PG x=(−600 km /h )+(−60.62 km /h )=−660.62 km /h

v⃑PG y= v⃑PW y+ v⃑WGy= (0km /h )+(−35 km /h )=−35 km /h

∑ v⃑PG=√ v⃑ PGx2+ v⃑ PGy

2=√ (−660.62 km /h )2+ (−35 km /h )2=660 km /h

tanθ=v⃑ PG y

v⃑PG x

∴ θ=tan−1( v⃑PGy

v⃑PGx)=tan−1( 35 km /h

660.62 km /h )=3.0 ° S of W

b) The pilot needs to travel due west. Calculate the new heading he needs to set for the airplane. Airspeed is still 600 km/h and the wind velocity is still 70 km/h 30° S of W.

Sine Law :sin B

b= sin C

c⇒ sin30 °

600 km /h= sinθ

70 km /h

Heading θ=sin−1[ (sin 30° ) (70 km /h )600 km /h ]=3.3 ° N of W

Relative Velocity in 2-D: Determine the velocity of Car A (object) relative to Car B (observer), v⃑AB.

v⃑AB= v⃑A⏟Object

− v⃑ B⏟Observer

¿ v⃑ AB= v⃑ A+(−v⃑ B)

v⃑AB=√ (v⃑ A )2+(− v⃑B)2=√ (35 km /h )2+(−15 km /h)2=38 km /h

tanθ=|v⃑B||⃑v A|

∴ θ=tan−1( 15 km /h35 km /h )=23 °

θ=Plane’ s Heading

v⃑AB=38 km /h , 23° S of E