PHYS 20 LESSONS

Unit 2: 2-D Kinematics

Projectiles

Lesson 3: Relative Velocity

Reading Segment #1:

Relative Velocity

To prepare for this section, please read:

Unit 2: p.11

C. Relative Velocity

The speed of an object is always in reference to some other

object.

For example, when we say an car is moving at 50 km/h,

this is in reference to the ground. The ground is assumed to

be at rest.

50 km/h

ground (at rest)

But what if the ground (i.e. the reference frame) is moving?

50 km/h

ground

10 km/h

In order to determine the velocity of an object relative to

a moving reference frame, we need to use vector arithmetic.

C1. 1-D Relative Velocity

We will first consider motion in one dimension.

Consider a person walking on a train:

4.0 m/s 10.0 m/s

The person is walking 4.0 m/s East (relative to the train),

while the train is moving 10.0 m/s (relative to the ground).

How fast is the person moving relative to the ground?

4.0 m/s 10.0 m/s

Since the person and the train are moving in the same direction,

the person appears to be moving even faster than the train.

vperson = (+4.0 m/s) + (+10.0 m/s) Ref: East +

= +14.0 m/s West -

= 14.0 m/s East

What if the person is walking in the opposite direction?

4.0 m/s 10.0 m/s

The person is walking 4.0 m/s West (relative to the train),

while the train is moving 10.0 m/s East.

How fast is the person moving relative to the ground?

4.0 m/s 10.0 m/s

The person and the train are moving in opposite directions.

But the train is moving faster.

vperson = (-4.0 m/s) + (+10.0 m/s) Ref: East +

= +6.0 m/s West

-

= 6.0 m/s East

Summary (1-D Relative Velocity)

If both objects are moving on the same axis:

State a reference system

- choose a positive and a negative direction

Simply add the vectors to get the relative velocity

Ex. 1 A boat, capable is travelling 9.0 m/s in still water, heads

East on a river with a current moving 4.0 m/s West.

a) What is its resultant velocity?

b) How long would it take to travel 6.0 km upstream?

a) current

4.0 m/s

9.0 m/s

Ref: East +

West -

a) current

4.0 m/s

9.0 m/s

Ref: East +

West - vboat = (+9.0 m/s) + (-4.0 m/s)

= +5.0 m/s

= 5.0 m/s East

b) v = d

t

d = v t

t = d = 6000 m

v 5.0 m/s

= 1.2 103 s

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 2 p. 13 #1, 2

Reading Segment #2:

2-D Relative Velocity

To prepare for this section, please read:

Unit 2: p.12

C2. 2-D Relative Velocity

We will now consider relative velocity in two dimensions.

This is especially useful for navigation.

Terminology:

Heading = The direction the plane is aimed

i.e. the way it would travel if there is no wind

Airspeed = The speed of the plane relative to the air

i.e. the speed it would travel if there was no wind

Groundspeed = The speed relative to the ground

i.e. the actual speed of the plane

There are two kinds of problems you will deal with:

1. Crosswind (or cross-current)

- the wind will blow you "off course"

2. Heading into the wind (or upstream)

- in order to get to a destination, you must

aim your craft into the wind

Ex. 1 (Crosswind question)

A plane heads directly North with an airspeed of 300 km/h.

However, there is a 50.0 km/h wind blowing from the East

(i.e. towards the West).

a) What is the resultant velocity of the plane?

b) How far would it get blown off-course in 40.0 minutes?

a)

300 km/h

The plane can travel 300 km/h if there is no wind.

a) 50.0 km/h

300 km/h

But, what happens to its velocity if there is

a 50 km/h crosswind?

a) N

50.0 km/h wind

R 300 km/h

plane

W E

Since the velocity vectors are at right angles (i.e. 90),add them tail-to-tip.

The resultant R represents the plane's actual velocity

(with respect to the ground).

a) 50.0 km/h

R 300 km/h

Pythag:

c2 = a2 + b2

R2 = (50.0 km/h)2 + (300 km/h)2

R = 304 km/h

Notice that the wind moves the plane even faster.

a) 50.0 km/h

R 300 km/h

Soh Cah Toa:

tan = 50.0

300

= tan -1 (0.1667) = 9.46

a) N

304 km/h

9.46

W E

So, the actual velocity of the plane (relative to the ground)

is

304 km/h at 9.46 W of N 80.5 N of W

99.5 rcs

b) First, find how far the plane travelled.

v = d

t

d = v t

= (304.138 km/h) (40 / 60 hours)

= (304.138 km/h) (0.66667 h)

= 202.76 km

b) N

202.76 km

9.46

W E

The plane's displacement is in the same direction

as its actual velocity.

Thus, it is directed at 9.46 W of N.

b) x

202.76 km

9.46

For a right triangle.

We are looking for the distance x.

b) x

202.76 km

9.46

Soh Cah Toa

sin = x

202.76 km

x = (202.76 km) (sin 9.46) = 33.3 km

Ex. 2 (Heading "into the wind")

A plane must fly directly West a distance of 1200 km.

However, there is a 65.0 km/h wind towards the North.

If the plane's airspeed is 340 km/h,

a) what heading is needed?

b) what is the flight time?

a)

Destination

340 km/h

If there was no wind, the plane could aim directly

West and travel at a speed of 340 km/h.

a)

Destination

However, if the plane aims West and there is a wind blowing

towards the North, then it will be blown off-course.

a)

Destination

The plane must aim "into the wind". Then, the wind will blow

the plane "on-course" and it will arrive at its destination.

a) v

60 km/h

(wind) 340 km/h

(still air)

a) v

60 km/h

(wind) 340 km/h

Soh Cah Toa: (still air)

sin = 60

340

= sin -1 (0.19118) = 11.0

a) W

11.0

340 km/h

S

So, the plane must head 11.0 S of W

(or 79.0 W of S, 191)

b) v

11.0 60 km/h

340 km/h

Find actual speed (groundspeed):

c2 = a2 + b2

3402 = v2 + 602

v2 = 3402 - 602

v = 333.73 km/h Notice that the wind slowed

the plane down.

b) Find time to travel 1200 km:

v = d

t

d = v t

t = d = 1200 km

v 333.73 km/h

= 3.60 h

Animations

Boat on Water:

http://mysite.verizon.net/vzeoacw1/velocity_composition.html

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 2 p. 13 #3 - 7