Lesson Notes By Weeks and Term - Senior Secondary School 1

SUBJECT: PHYSICS

CLASS:  SS 1

DATE:

TERM: 1st TERM

REFERENCE BOOKS

• New School Physics. By Prof. M.W Anyakoha 
• New System Physics. By Dr. Charles Chow et.al

 
WEEK FIVE AND SIX

TOPIC: VECTOR and SCALAR QUANTITY, DISTANCE/DISPLACEMENT, SPEED/VELOCITY, ACCELERATION, DISTANCE/DISPLACEMENT - TIME GRAPH, SPEED/VELOCITY - TIME GRAPH

CONTENT

• Scalar & Vector Quantity
• Distance & Displacement
• Speed & Velocity
• Acceleration & Retardation
• Distance/Displacement - Time Graph
• Speed/Velocity - Time Graph

FOR WEEKS FIVE AND SIX

SCALAR & VECTOR QUANTITY

A scalar quantity is defined as a quantity that has magnitude only but no direction. Typical examples of scalar quantities are time, distance, speed, temperature, volume, work, power, electric potential etc.  A scalar quantity or parameter has no directional component, only magnitude.  For example, the units for time (minutes, days, hours, etc.) represent an amount of time only and tell nothing of direction.  Additional examples of scalar quantities are density, mass, and energy.

A vector quantity is defined as a quantity that has both magnitude and direction. Typical examples of vector quantities are velocity, displacement, acceleration, force, momentum, moment, electric field intensity etc

POSITION

Position is referred to as the point in which an object can be located or the place object is found. The position of an object on a plane can be given by its co-ordinates, i.e., the signed distances of the point from two perpendicular axes, OX and OY

        Y

                                            X       

Fig. 6.0 Cartesian co-ordinates

The – co-ordinates is called abscissa while the – co-ordinate is called ordinate. The co-ordinate is written first, before the – co-ordinates, i.e. (X,Y) 

DISTANCE AND DISPLACEMENT

Distance: This is the gap between any two positions in space. It is denoted by S and measured in metre(m)  it is a scalar quantity and is calculated as the product of average speed and time.

Thus, distance = average speed X time.  

Displacement: This is the distance covered in a specific direction. it is a vector quantity measured in metre(m). The direction of motion of bodies can be found by using the compass.

Displacement = average velocity X time. It is denoted by X 

The Use of Bearing to Indicate Direction and Displacement

The bearing of an object from the origin is the angle which it makes with the north pole in the clockwise sense. It is specified in two ways:

1. The use of cardinal points: N – North, S – South, W – West, and E – East
2. The use of three digit notation. Students should note that bearing which are located by cardinal points are with respect or reference to the North and South.

Fig. 6.1 cardinal points and their directions 

SPEED AND VELOCITY

Speed: Speed is defined as the rate of change of distance moved in an unspecified direction or the rate of change of distance per unit time in an unspecified direction. It is measured in metre per second (m/s). It  is a scalar quantity.

The mathematical expression of speed is 

Average Speed: Average speed is defined as the ratio of the total distance travelled to the total time taken. It is a scalar quantity and measured in m/s or ms-1

This, average speed =

When a body covers equal distance in equal time intervals, no matter how small the time interval may be, it is said to be a uniform speed or constant speed.

Velocity: Velocity is defined as the rate of change of distance moved in a specific direction or the rate of change of displacement. Velocity is a vector quantity. For instance, it would be easy and correct to say that a car travelling at a steady speed of 50km/h in a direction of N40oE has a velocity of 50km/h, N40oE.

 velocity =    

Uniform velocity

Fig 6.2 Uniform Velocity

Uniform (constant) velocity: An object is said to undergo (constant) velocity, if the rate of change of displacement is constant, no matter how small the interval may be.

Example 1:

A train moves with a speed of 54km/h for one quarter minute. Find the distance travelled by the train.

Solution:

    Speed    =    54km/h    =   15m/s

    Time    =    ¼ min     =    ¼ × 60  =  15s

    Distance     =    speed (m/s)  ×  time (s)

              =    15(m/s)  ×  15(s)       

       =    225m

ACCELERATION & RETARDATION

Acceleration is defined as the increasing rate of change of velocity. It is measured in m/s2.

Acceleration (a) = Increasing Velocity change

                      Time taken                .   ……………………………………5.

When the velocity of a moving body increases by equal amount in equal intervals of time, no matter how small the time intervals may be, it is said to move with uniform acceleration.

Retardation is defined as the decreasing rate of change of velocity. It is measured in m/s2.It is also known as deceleration or negative acceleration

Retardation (ar) = Decreasing Velocity Change

                               Time Taken               

EQUATION OF UNIFORMLY ACCELERATED MOTION

S = (v+u) t                        ………………………………………………………7

          2

v = u + at                        ……………………………………………………….8

v2 = u2 + 2 aS                    ……………………………………………………….9

S = ut + ½ at2                   ……………………………………………………….10

Equations (7) to (10) are called equations of uniformly accelerated motion and could be used to solve problems associated with uniformly accelerated motion

 where u- initial velocity( m/s), v – final velocity (m/s), a – acceleration (m/s2), s – distance covered and t – time (m).  

Example 2

A car moves from rest with an acceleration of 0.2mls2 . Find its velocity when it has moved a distance of 50m.

Solution:

a = 0.2mls2 , S = 50m, u = 0m/s , v = ?

v2 = u2 + 2 as

v2 = 02 + (2x0.2x50) = 20

v = √20 m/s

EVALUATION

1. State the differences & similarity between speed & velocity.                                                              2.          A car has a uniform velocity of 108km/hr. How far does it travel in ½ minute?

GRAPHS

The motion of an object is best represented or described with graphs. These graphs are 

1. Distance- time
2. Displacement – time
3. Velocity – time

Distance – time

In a distance-time graph, its slope or gradient gives the speed.

    (i) Uniform speed                    (ii) Non-uniform speed

Fig. 6: Distance-time graph

Gradient/slope    =    speed   =   

Displacement – time graph

A displacement-time graph could be linear or curved. For a linear graph, the gradient gives the velocity.

1. a) Non-uniform velocity

Fig. 6.4 Displacement-time graph

Gradients/slope    =  velocity (v)  =    

Velocity – time graph

The velocity-time graph is more useful than any of the two graphs described above because it gives more useful information concerning the motion of objects. The following information can be obtained from the graphs (i) acceleration (ii) retardation  (iii) distance   (iv) average speed.

The motion of objects can form shapes such as square, triangle, trapezium, rectangle or a combination of two or more shapes. Thus, the sum of the areas of the shapes formed corresponds to the distance moved, covered or travelled by the objects.

Example 3

A motor car accelerates for 10secs to attain a velocity of 20m/s. It continues with uniform velocity for a further 20 seconds and then decelerates so that it stops in 20 seconds. Calculate (i) Acceleration  (ii) Deceleration  (iii) The distance travelled.

1. i)     or   

20    =   

A     =   

1. ii) Deceleration =

iii)    Using area of trapezium

    ½ × (AB + OC) h      =     ½ × (20 + 50) 20       

        =    ½ × (70) × 20           =     700m

Example 4

A car starts from rest and accelerates uniformly until it reaches a velocity of 30mls after 5 seconds.  It travels with uniform velocity for 15 seconds and is then brought to rest in 10s with a uniform retardation. Determine (a) the acceleration of the car (b) The retardation  (c) The distance covered after 5s (d) The total distance covered (use both graphical and analytical method).

The velocity – time diagram for the journey is shown above, from this diagram

1.     the acceleration = slope of OA

= AE / EO

= (30-0) /(5-0)=30/5

= 6mls2 

1.     the retardation = slope of BC = CB / CD

= (0-30) / (30-20) = -30/10

= -3mls2  (the negative sign indicate that the body is retarding)

1.   Distance traveled after 5s   = area of A E O

                = ½ x b x h

                = ½ x 5 x 30

                = 75m

1. Total distance covered  = area of the trapezium OABC

= ½ (AB + OC) AE

= ½ (15 + 30) 30

= 675m.

Using equations of motion.

1. U = O, V = 3, t = 5

V = u + t

a = v-u/t = 30 – 0 / 5

a = 30/5 = 6ms-2 

1. a o in 

a = v – u / t   = 0-30 / 10

a = -3 mls2 

  (c) S = ( u + v)  5

         2

        = 30 / 2 x 5

        = 75m

 (d) To determine the total distance travelled,  we need to find the various distance for the three stages of the journey and then add them.

for the 1st part         S= 75m  from (c)

for the 2nd stage where it moves with uniform velocity.

            S = vt

            = 30 x 15

            = 450m

for the last stage S = ½ (u + v) t

            = ½ (30 + 0) 10

            = 150m.

                   Total distance = 75 + 450 + 100 = 675m.

EVALUATION

1. A train slows from 108km/hr with uniform retardation of 5mls2. How long will it take to reach 18km/hr and what is the distance covered?.
2. Why is velocity – time more useful than displacement time graph?

READING ASSIGNMENT

www.google.com (click on google search, type “ distance & displacement ”, click on search) & New school physics by M.W.Anyakoha,Ph D  Pg 14 – 18

WEEKEND ASSIGNMENT

1. A body which is uniformly retarded comes to rest in 10s after travelling a distance of 20m. Calculate its initial velocity (a) 0.5 ms−1 (b) 2.0ms−1  (c) 4.0ms−1     (d) 20.0 ms−1  (e) 200.0 ms−1
2. The distance travelled by a particle starting from rest is plotted against the square of the time elapsed from the commencement of the motion. The resulting graph is linear. The sped of the graph is a measure of   (a) initial displacement  (b) initial velocity  (c) acceleration (d) speed  
3. Which is the in correct formula for a body accelerating for a body accelerating uniformly? (a) (b) (c)

(d)     (e)

4. The slope of a displacement-time graph is equal to 

    (a) acceleration   (b) uniformly velocity   (c) uniform speed  (d) instantaneous speed

5. A body moving with uniform acceleration has two points (5, 15) and (20, 60) on the velocity-time graph of its motion. Calculate (a) 0.25 ms−2 (b) 3.00 ms−2 (c) 4.00 ms−2 (d) 9.00ms−2
6. A moving object is said to have uniform acceleration if its (a) displacement decreases at a constant rate  (b) speed is directly proportional to time (c) velocity increases by equal amount in equal time intervals (d) velocity varies inversely with time 
7. The diagram shows a velocity-time graph of the motion of a car. What is the total distance covered after the journey? (a) 75m (b) 150m (c) 300m (d) 375m
8. The area under a velocity-time graph represents (a) final velocity attained     (b) direct covered    (c) acceleration    (D) workdone
9. A body accelerators uniformly from rest at 2ms−2. Calculate its velocity after travelling 9m. (a) 36 ms−1    (b) 18 ms−1    (c) 6 ms−1    (d) 4.5 ms−1
10. A moving object is said to have uniform acceleration if its 

    (a) displacement decreases at a constant rate    (b) speed is directly proportional to time (c) velocity increases by equal amount in equal time intervals    (d) velocity varies inversely with time

THEORY

1. A body moving with uniform acceleration a, has two points (5, 15) and (20, 60) on the velocity-time graph of its motion. Calculate the acceleration a.
2. Two points on a velocity- time graph coordinates (5s, 10ms-1) and (20s, 20ms-1). Calculate the mean acceleration between the two points.
3. A car starts from rest and accelerates uniformly for 5s until it attains a velocity of 30ms-1. It then travels with uniform velocity for 15s before decelerating uniformly to rest in 10s;

    (i)  Sketch a graph of the motion

    (ii) Using the graph above, calculate the 

    (a) Acceleration during the first 5s

    (b) Deceleration during the last 10s

    (c) Total distance covered through the motion

4. A car starts from rest and accelerates uniformly for 10s, until it attains a velocity of 25m/s, it then travels with uniform velocity for 20s before decelerating uniformly to rest in 5s.

(i) Calculate the deceleration during the last 5s

(ii) Calculate the acceleration during the first 10s

(iii) Sketch a graph of the motion and calculate the total distance covered throughout the motion.

5. (a) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph

    (i) Acceleration        (ii) Retardation    (iii) Total distance

(b) Show that the displacement of a body moving with uniform acceleration a is given by S = ut + 1/2at2, where u is the velocity of the body at time t=0

(c) A particle moving in a straight line with uniform deceleration has a velocity of  40m/s at a point P, 20m/s at a point Q and comes to rest at a point R, where QR=50m. Calculate the: 

    (i) Distance PQ (ii) Time taken to cover PQ (iii) Time taken to cover PR (WAEC, 1990)

6. (a) What is meant by the statement the acceleration of free fall due to gravity on the equator is 9.78ms-2

    (b) State two factors that affect the value of the acceleration due to gravity.(WAEC,2006)

7. Using suitable diagram, explain how the following can be obtained from a velocity- time graph: (a)Acceleration (b) Total distance covered      (c) A body at rest is given an initial uniform acceleration of 6.0ms-2 for 20s after which the acceleration is reduced to 4.0ms-2 for the next 10s. The body maintains the speed attained for 30s. Draw the velocity-time graph of the motion using the information provided above. From the graph, calculate the:

(i) Maximum speed attained during the motion

(ii) Total distance travelled during the first 30s

(iii) Average speed during the same time interval as in (ii) above (WAEC, 2009)

8. (a) Sketch a distance-time graph for a particle moving  in a straight line:

    (i) Uniform speed (ii) Variable speed (NECO, 1010)

    (b) A body starts from rest and travels distances of 120, 300, and 800m in successive equal     time intervals of 12s. During each interval the body is uniformly accelerated.

    (i) Calculate the velocity of the body at the end of each successive interval.

    (ii) Sketch the velocity- time graph of the motion. (WAEC, 2010)

9. (a) Explain the terms: uniform acceleration and average speed. 

(b) A body at rest is given an initial uniform acceleration of 8.0ms-2 for 30s after which the acceleration is reduced to 5.0ms-1 for the next 20s. The body maintained the speed attained for 60s after which it is brought to rest in 20s.Draw the velocity-time graph of the motion using the information given above.

    (c) Using the graph, calculate the:

    (i) Maximum speed during the motion.

    (ii) Average retardation as the body is being brought to rest.

    (iii) Total distance travelled during the first 50s.

    (iv) Average speed during the same interval as in (ii) above ( WAEC, 1991)

10. (a) State two reasons why the acceleration due to gravity varies on the surface of the earth. (NECO, 2008)

(b) State the difference between centripetal and centrifugal force.(NECO, 2011)

11. (i) Define velocity and acceleration

    (ii)List two physical quantities that can be deduced from a velocity-time graph.

              Define the following  terms; (a) average speed (b) Instantaneous velocity

(c) A car travels at an average speed of 20ms-1. Calculate the distance covered in 1hour

1. Starting from rest, a vehicle accelerates at 2m/s2 for 5secs it then travels for 5secs at the velocity, Vo reached and is brought to rest with a uniform retardation after the next 5s.

1. Sketch the velocity-time graph for the journey
2. Calculate the value of Vo,
3. What is the retardation 
4. The total distance covered