Derivation of Equations of Linear Motions SS2 Physics Lesson Note

BASIC DEFINITIONS

1. Displacement: This is the distance travelled in a specified direction. It is a vector quantity. Its unit is meters
2. Distance: This is the space or separation between two points. It is a scalar quantity. Its unit is meters
3. Speed: this is the rate of change of distance with time. It is a scalar quantity. Its unit is a meter per second (m/s)

Speed= distance

    Time

4. Velocity: this is the rate of change of distance with displacement with time. It is a vector quantity. Its unit is a meter per second (m/s)

Velocity= displacement

Time

5. Acceleration: this is the increasing rate of change of distance with time. It is a vector quantity. Its unit is meter per seconds-square (m/s2). Retardation or deceleration is a negative acceleration. 

Acceleration=  velocity

                         Time

DERIVATION OF EQUATIONS OF LINEAR MOTION

v= final velocity 

u = initial velocity 

a = acceleration 

t = time 

s = distance

Average speed = total distance

Time

Total distance= average speed x time

s= (u + v) x t ————– (1)

         2

From the definition of acceleration

a = (v-u) ————– (2)

t

From equation (2) substitute for ‘t’ into equation (1)

v2 = u2 + 2as ————– (3)

From equation (2) substitute for ‘v’ into equation (1)

s = ut + ½(at2) ————– (4)

Calculations Using the Equation of Motion

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

u= 0m/s; a= 0.2m/s2; s= 50m; v =? 

v2 =u2 + 2as 

v2 = (0)2 + 2 (0.2 x 50) 

v2 = 20

v = √20

v = 2√5m/s 

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

Solution

v=108 km/hr; t= ½ minutes; Distance =? 

v = 108 km/hr = 108 x1000

                            3600

v= 30 m/s

t= ½ 60 = 30secs 

Speed = distance

time

Distance = speed x time

s = 30 x 30

s = 900 m

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, it must be noted that:

a= g is positive for a downward motion 

a= -g 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 re-bouncing body, the height 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; initial velocity of an object dropping from a height) 

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.

Example:

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 x 20 /10)

t = 2secs 

v = u + gt

v= gt

v = 10 x2 

 v = 20 m/s 

ASSIGNMENT 

1. A ball thrown vertically upwards from the ground level hits the ground after 4s. Calculate the maximum height reached during its journey
2. A particle starts from rest and moves with a constant acceleration of 0.5m/s2. Calculate the time taken by the particle to cover a distance of 25m.
3. A car takes off from rest and covers a distance of 80m on a straight road in 10 seconds. Calculate its acceleration
4. A particle accelerates uniformly from rest at 6m/s2 for 8 secs and then decelerates uniformly to rest in the next 7 secs. Determine the magnitude of the deceleration.