Lesson Notes By Weeks and Term - Senior Secondary School 2

SUBJECT: PHYSICS

CLASS:  SS 2

DATE:

TERM: 1st TERM

REFERENCE TEXT

• New School Physics by M.W Anyakoha
• SSCE WAEC Past Questions
• UTME Past Questions

 
WEEK THREE

TOPIC: DERIVATION OF EQUATONS OF LINEAR MOTION

CONTENT

• Basic definitions
• Derivation of equations of linear motion
• Motion under gravity

Basic definitions

There are four major terms associated with motion in a straight line . These are speed (v) or velocity (v), distance (s)  or displacement (s), acceleration (a) and time (t).

DISPLACEMENT: This is the distance traveled in a specified direction. For example, if a body moves a distance of 50m northwards, it is a vector quantity   while distance is a scalar quantity.

Distance indicates how far an object has moved. It’s  a scalar quantity.

The rate at which a body covers a distance is called the SPEED of the body.

Thus, speed = distance 

                      Time              ( m/s   , km/hr)

VELOCITY is the rate of change of displacement with time. When a body moves with equal displacement in equal interval of time, no matter how small the time intervals may be, the velocity is said to beuniform or constant. 

ACCELERATION: is the rate of change of velocity with time . When  the velocity  increases  in time , the rate of change of  velocity  is termed acceleration  but when  the velocity  decreases with time  ,it is  called RETARDATION .Retardation is a negative acceleration. 

When the rate of change of velocity with time is constant, we have uniform  

acceleration

EVALUATION

Sketch the velocity—time graph for a body  that starts from rest and accelerates uniformly  to a certain velocity If it maintains this for  a given period before its eventual deceleration. Indicate the following: 

             1 Uniform acceleration, retardation

             2 Total distance travelled 

DERIVATION OF EQUATIONS OF LINEAR MOTION

v= Final velocity 

u = Initial velocity 

a = Acceleration 

t = Time 

s = Distance 

v= u + at      --------------------- (i)

Average speed = total distance

Time

s = u+v

t     2

   s = u+u+at

t      2 

     s =  2u+ at 

t        2

s   = u + ½ at

t

     s= ut + ½ at2 ________________(2)

     v2 = (u + at )2

 v2 = (u + at)(u + at) 

v2 = u2 +2uat + at2

=  u2 +  2a (ut + ½at2)

therefore   v2 = u2 +2as---------------(3) 

                s= ( u + v ) t --------------(4)

                         2

CALCULATIONS USING THE EQUATION OF MOTION

A car moves from rest with an acceleration of 0.2 m/s2. Find its velocity when it has covered distance of 50m 

u= 0m/s

a= 0.2m/s2

s= 50m

v = ?

v2 =u2 + 2as    = (0)2 + 2 ( 0.2 x 50)  = 20

v= √20    = 2√5m/s 

A car travels with a uniform  velocity of 108km/hr .How far does it travels in  ½  minutes?

Solution

v=108km/hr      t= ½  minutes      distance = ? 

v = 108 km/hr = 108 x1000

                            3600

v= 30m/s        t= ½ 60 = 30secs 

speed = distance 

time

 Distance = speed x time   = 30 x 30    = 900 m

CLASS ACTIVITIES

(1)  A train slopes from 108 km/hr with a uniform retardation of 5 m/s2 . How long will  it take to reach  18 km/hr   and what is the distance  covered ? 

(2)  An orange fruit drops to the ground  from  the top  of  a tree 45m tall .How long does it take to reach the ground (g= 10m/s2)? 

(3)   A car moving with a speed of 90 km/h was brought uniformly to rest by the application of brake in 10s. How far did the car travel after the far did the car travel after the brakes were applied.Calculate the distance it covers in the last one second its motion.

FURTHER ACTIVITIES

A car starts from rest and accelerates uniformly until it reaches a velocity of 30m/s  after 5secs . It travels with this uniform velocity for 15secs and it is then brought to rest in 10 secs with a uniform acceleration. Determine 

(a)           the acceleration of the car 

(b)          the retardation 

(c)          the distance  covered  after 5 secs

(d)         the total  distance  covered. 

SOLUTION 

V (m/s)

                             A                                B 

     30                         

                            E                             D             C                                               

       O                  5                                20          30 

ACCELERATION =   AE=  30 =   6m/s2

                          EO       5

RETARDATION  =CB  =   0 - 30   = - 3m/s2

                         DC       10       

II The distance covered  after  5sesc 

The area  is given  by area  of the triangle 

                           = ½ b h  

=  ½  (5) 30 

                            = 75m

iv   The total distance  covered = area of the trapezium OABC

                                  = ½ (AB + OC ) X h

                                   = ½ (15 + 30 ) 30 

                                   = 45 x 15 

                                   = 675 m

CLASS WORK 

A lorry starts from rest and accelerates uniformly until it reaches a velocity  of 50 m/s  after  10 secs . It travels with uniform velocity for 15 secs  and is brought to rest I 5secs  with a uniform  retardation .

Calculate :

1. a)  The acceleration of the lorry 
2. b)  The retardation 
3. c)   The total  distance  covered 
4. d)  The  average  speed  of the lorry

MOTION UNDER GRAVITY

A  body  moving  with a uniform acceleration  in space does so  under  the influence  of gravity  with a constant  acceleration . (g = 10 m/s2 ). In dealing with vertical motion under gravity , the following  points must be noted

• All objects dropped near the surface of the Earth in the absence of air resistance fall toward the Earth with the same nearly constant accelerationWe denote the magnitude of free-fall acceleration as g.
• The magnitude of free-fall acceleration decreases with increasing altitude. Furthermore, slight variations occur with latitude. At the surface of the Earth the magnitude is approximately 9.8 m/s². The vector is directed downward toward the centre of the Earth.
• Free-fall acceleration is an important example of straight-line motion with constant acceleration.
• When air resistance is negligible, even a feather and an apple fall with the same acceleration, regardless of their masses.

• a= +g is positive  for a downward motion but (-g)  negative for  an upward  motion 
• The velocity v = 0  at maximum height  for a vertically projected object
• The initial velocity u = 0 for a body dropped from rest above the ground .
• For a rebounding body the height  h  above the ground is zero 

The time of fall of two objects of different masses has nothing to do with their masses   but is dependent  on the distance  and acceleration  due to gravity as  shown  below

S= ut  + ½ gt2

S = ½ gt2 (u=0) 

 t=   √2s/g 

The above relationship can also   be used to determine the value of acceleration due to gravity. If we plot s against t, it will give us a parabolic curve.

S(m)

        parabol

     t (s)

A graph of s against t2  will give us  a straight line through the   origin  with slope  ½ g 

from which  g  can  be  computed           

S(m) 

         Slope  = ½ g 

       (O,0)                                          

t2 (s2)

Case One. For a body projected from a tower or plane of height h.

1. The body covers both horizontal Sx( also known as the range , R) and vertical Sy ( height) distance.

Gravity has no effect on the horizontal distance covered but on the vertical distance, hence

Sx = R= ut……………………  #

Sy = ut + 1/2gt2    but u = 0

Sy = ½ gt2 ……………………. #

Case two: For a body thrown vertically upward from the ground to a maximum height h and back to the ground. 

At maximum height v = 0

Time taken to maximum height is same time taken from maximum height to the ground.

 Time to maximum height (t) 

V = u – gt

0 = u – gt

u = gt…………….. # 

Maximum height attained

V2 = u2 - 2ghmax

0 = u2 – 2ghmax

u2 =  2ghmax …………………..  #

Case three for a body projected from the top of a tower to a maximum height h

At maximum height h, v=0

Time (t) to maximum height 

V = u - gt

0 = u –gt

U = gt

Time from maximum height h to the ground 

Total distance travelled = h + h1

h + h1 = ut + 1/2gt2

CALCULATIONS

1. A ball is thrown vertically into the air with an initial velocity, u. What is the greatest height reached? 

Solution  

V2 = u2+  2as

U= u  , a = -g   ,  v = 0 

02 = u2 + 2 (-g) s

2gs = u2

s = u2 /2g 

2. A ball is released  from a height  of 20m .Calculate 

 (i) the time it takes to fall 

(ii) the velocity  with which it hits  the ground 

a= +g   u=0     s =20m    t = ?   

t  = √2s/g  

t = √ 2 x20 /10 

t = 2 secs

v = u + gt

v= gt

v = 10 x2 

v = 20 m/s 

GENERAL EVALUATION

1. List five apparatus for measuring the mass of a body.
2. List five apparatus for measuring the length of a body.

WEEKEND ASSIGNMENT

1. A body starts from rest and accelerate  uniformly at 5m/s2 until it attain a velocity of 25m/s.Calculate the time taken to  attain this velocity (A) 2S (b) 3s (c) 5s (d) 6s.
2. A particle accelerates uniformly from rest at 6m/s2 for 8secs and then decelerates uniformly to rest in the next 5 secs. Determine magnitude of the deceleration

(a) 9.6 m/s2 (b) -9.6 m/s2 (c) 6.9 m/s2 (d) – 6.9 m/s2

3. A car takes off  from rest and covers a distance of 80m on a straight road in10secs  Calculate its acceleration (a) 160 m/s2 (b) 16 m/s2  (c) 1.6 m/s2 (d) 0.16 m/s2
4. An object is released from rest at a height of 20m. Calculate the time it takes  to fall to the ground ( g= 10m/s2) (a) 1s (b) 2s (c) 3s (d) 4s. 
5. A body accelerates uniformly rest at the rate of 3m/s2 for 8 secs. Calculate the distance it covers. (a) 24m (b) 48m (c) 72m (d) 96m.

THEORY

1. A particle start from rest and moves with constant acceleration of 0.5m/s2 .Calculate  the time taken by the particle to cover a distance of 25m . 
2. A particle accelerate uniformly from rest at 6m/s2 for 8secs and then decelerates uniformly to rest in the next 7secs .Determine the magnitude of the deceleration.

READING ASSIGNMENT

New Sch. Physics for Senior Sec. Schls. Pages 130-134.