Displacement/Relative displacement

 

FURTHER MATHEMATICS SS 1 SECOND TERM

 

 

 

WEEK 6 DATE(S)……………………………………….

SUBJECT: FURTHER MATHEMATICS

CLASS: SS1

TOPIC: RELATIVE VELOCITY

CONTENT:

1. Displacement/Relative displacement
2. Acceleration
3. Relative velocity
4. Motion of word problem

SUB – TOPIC: Relative velocity

Every object or occurrence (be it real or abstract) is relative to some other object, occurrence. If a ship has a speed of 10m/s in still water, then we are referring to its speed relative to earth.

Note the following points when solving relative velocity problems:

1. Identify the parameters in the problem i.e. relative velocity, magnitude, direction.
2. Draw the displacement diagram.
3. Write the relative velocity equation, making use of the Domino rule; e.g. _AV_B =_AV_E+_EV_B i.e. the velocity of A relative to B is equal to the velocity of A relative to the earth plus velocity of earth relative to B. This rule can be extended to any number of terms.

Example 1: Ship A is sailing due East at 12m/s and ship B is sailing due North at 16m/s. Find the velocity of ship B relative to ship A.

Solution:

Magnitude Direction

_AV_E 12m/s East

_BV_E 16m/s North

Since _BV_E = _BV_A + _AV_E and we seek _BV_A, then _BV_A = _BV_E – _AV_E

Displacement diagram:

E

A

_BV_E

_BV_A

_AV_E

 

 

B

Applying Pythagoras’ theorem to the triangle above gives:

_BV_A = , and makes direction with _BV_E

Example 2: If ship A is moving North – east at 15m/s and a second ship B appears to an observer from A to be moving east at 7m/s. Find the actual velocity of B.

Solution: Magnitude Direction

_AV_E 15m/s North – East

_BV_A 7m/s East

Displacement diagram

_BV_E

_AV_E

_BV_A

135

Since _BV_E = _BV_A + _AV_E

Applying cosine rule, (_BV_E)² = 7² + 15² – 2 × 7 × 15

Therefore, _BV_E =20.6m/s

Applying sine rule,

Therefore

Therefore ⁰

The direction is about N59⁰E

CLASS ACTIVITY

1. One ship is sailing due east at 24km/h and another ship is sailing due north at 32km/h. Find the velocity of the second ship relative to the first.
2. If a ship A is moving North – East at 12km/h and an engine boat B appear to an observer from A to be moving south at 10km/h. Find the actual velocity of ship B.

SUB – TOPIC: Displacement

Displacement is the distance moved by a point in a specified direction. It is a vector quantity, and therefore has both magnitude and direction. If I walk 4km East and 3km South, then I have been displaced 5km from my starting point, despite the fact that I have walked 7km altogether.

5km

3km

4km

The displacement of a point moving from the point A to the point B can be with respect to the origin O , can be represented by the vector AB.

If the position vectors of A and B are a and b respectively, then:

AB = AO + OB

= – OA + OB

= – a + b

Relative displacement

R_t

P_t

R_O

P_O

_tV_R

_tV_p

We use the diagram above to consider the displacement vector of P relative to R. Let initial position of P be P_O and that of R be R_O. Let the position of P after time t be P_t, and that of R after time t be R_t. Let the velocity of P be V_P and the velocity of R be V_R. After a time t, P will be at P_t so

OP_t = OP_O + tV_P

Similarly, OR_t = OR_O + tV_R

Considering the vector polygon above,

R_tP_t = R_tR_o+ R_OP_O + P_OP_t

= – _tV_R + R_OP_O + tV_p

= R_OP_O + _tV_P – _tV_R

= R_OP_O + t(V_P – V_R)

Hence: R_tP_t = R_OP_O+ t(V_P – V_R)

i.e. displacement = time × Velocity ( of P relative to R)

Example 1: An aircraft A flies at 800km/h due west. Another aircraft B, which is 20km due north of A sets off to pursue A at 1200km/h. Find:

1. The course of aircraft
2. The time of interception of the aircraft A by the aircraft B

Solution: the velocity diagram is shown below

_AV_E

_BV_A

B

A

_BV_E

Let _BV_E = velocity of B relative to earth

_AV_E = velocity of A relative to earth

_BV_A = velocity of B relative to A

_BV_E

Then, ₌ 1200

_AV_E

= 800

=

to the nearest whole number.

Hence, the course (direction) of aircraft B is (i.e. .

(ii) to get the time of interception, we consider the displacement diagram

A

B

20

Let the distance along the course be , then

; i.e.

Solving for we get

time =

=

=

= 0.0224hr

Example 2: A boat A sails due west at 15km/h. Another boat B, initially 40km due north of the first, sail at 10km/h on the bearing. Find:

1. The shortest distance between the two boats
2. The time when they are at this distance.

Solution: consider the velocity diagram below

-V_B

V_A – V_B

V_A

P

Q

R

Where V_A = velocity of boat A

V_B = velocity of boat B

V_A – V_B = velocity of boat A relative to boat B

and

Let . Then using the cosine rule;

 

= 225 + 100 –

 

Let then using the sine rule,

;

 

Let be the initial position of boat A

be the initial position of boat B

be the position of boat B at time t

Consider the displacement diagram below

The shortest distance between the two boats is given by the length of the perpendicular from to Let the perpendicular from to be

Then:

.

 

 

Since and

Then

=

 

 

= 29.85km

(ii) Let t be the time that they are at the shortest distance apart, then

 

=

=

=

= 2.51hr

CLASS ACTIVITY

1. A helicopter travels at 100km/h in still air. If the wind is blowing from the west at 40km/h, how long will it take the helicopter to reach a place 250km off to the south – west?
2. A trawler A is 40km west of another traveler B. A set off at 20km/h on a course of 060⁰. If the trawler B can travel at 25km/h, what course should trawler B take to intercept trawler A?

SUB – TOPIC: Acceleration/motion of Aircraft and Ship

Acceleration is the rate of change of velocity with time. It is a vector quantity. If the magnitudes of accelerations represented by the adjacent sides of a parallelogram are and , with the angle between them being β, then the magnitude of the resultant acceleration denoted by is given by:

. If = 90⁰, then

When an aircraft moves through the air, it is affected by the motion of the air. The diagram below illustrates the relationship between the course/airspeed, direction of wind/wind speed and track/ground speed.

Wind direction/wind speed

Course/airspeed

Track/ground speed

The angle of drift is . The motion of a ship in water is similar to the motion of an aircraft in the air.A similar vector diagram for the motion of a ship in water is shown below.

Current speed

Water speed

Speed of ship

Example 1: an aircraft flies with an air speed of 160km/h from its base P to a point Q which is 65km East and 130km North of P. The wind is blowing from the west at a speed of 32km/h. find the:

1. Direction in which it is headed
2. Time, to the nearest minuet, taken on its journey from P to Q

Solution:

R

P

θ

α

N

Ground speed

Wind speed

β

Air speed

Q

T

130km

α

65km

 

 

From ,

= 2

α = 63.43⁰

1

2

α

Given that

It follows that

Wind speed = 32km/h

Air speed = 160km/h

Using sine rule in

 

 

=

=

 

 

=

= 16.26⁰

Hence the direction in which the aircraft is headed is N16.26⁰E

160km/h

32km/h

V

α

θ

β

Let the ground speed be V, then using sine rule,

 

=

=

We get From above thus;

 

 

=

= 106.26

 

= 171.7km/h

130km

65km

P

R

Q

Consider the displacement diagram PRQ below:

In , let y be the distance from P to Q, then

 

= 145.3

Let the time taken to fly from P to Q be t

Since time =

 

=

= 0.8462h or 51mins.

CLASS ACTIVITY:

1. An aircraft flew out of Benin Airport on a bearing of 315⁰ and at an air speed of 250km/h. if the speed and direction of the wind were 50km/h and 075⁰, calculate, correct to the nearest whole number:
2. The ground speed
3. The angle of drift
4. The track
5. A ship P is sailing due east with a speed of 4µm/s. The speed of the second ship Q relative to P is µ m/s south – west. Find in m/s, the speed of Q.

PRACTICE QUESTIONS

1. A ship P sails due east with speed 4tm/s. A second ship Q sails south – west with a speed of m/s relative to Q. Find in m/s, the speed of Q.
2. Two particles A and B have velocities and respectively. Find the velocity of B relative to A.
3. An aircraft has airspeed of 96km/h and flies in the direction S30⁰W. A wind blows 40km/h from the direction N40⁰W. Calculate the ground speed of the aircraft.
4. The airspeed of an aircraft is 250km/h and the wind blows from the direction S45⁰W with a speed of 40km/h. the pilot of the aircraft wishes to travel due north. In what direction must he steer the aircraft?
5. Two cars A and B are travelling in perpendicular directions along straight roads which intersect at O. A moves with a velocity of 30km/h and B moves with a velocity of 50km/h. Find the velocity of A relative to B and the direction of car A.

ASSIGNMENT

1. The position vector of a point P is and the point Q has position vector where are unit vectors in direction of East and North respectively. Calculate:
2. The distance PQ, correct to the nearest 0.1m
3. The bearing of Q from P, correct to the nearest degree.
4. A and B are two airports with B 400km due east of A. An aircraft whose speed in still air is 200km/h flies directly from A to B through a wind blowing from the northeast at 40km/h. Find the speed of the aircraft relative to the ground. On another day, the aircraft set out from A to B in still air. At the same time another aircraft starts to fly north from B at 150km/h. Determine the shortest distance between the two aircrafts and the time taken, correct to the nearest minute, before they are closest to each other.

KEY WORDS

• VELOCITY
• DISPLACEMENT
• ACCELERATION
• COURSE
• AIRSPEED
• GROUND SPEED
• ANGLE OF DRIFT
• BEARING
• TRACK