# DERIVATION OF EQUATONS OF LINEAR MOTION

## BASIC DEFINITIONS

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

Speed= distance

Time

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

Velocity= displacement

Time

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

Acceleration= velocity

Time

EVALUATION I

Sketch the velocity-time graph for a body that starts from rest and accelerates uniformly to a certain velocity. If it maintains this forÂ  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

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)Â Â Â Â Â  ————– (2)

t

From equation (2) substitute for â€˜tâ€™ into equation (1)

v2 = u2 + 2asÂ Â Â Â Â Â Â Â  ————– (3)

From equation (2) substitute for â€˜vâ€™ into equation (1)

s = ut + Â½(at2) ————– (4)

Â

## 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 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 = âˆš20

v = 2âˆš5m/s

1. A car travels with a uniform velocity of 108km/hr .How far does it travels in Â½ a minute?

Solution

v=108km/hr; t= Â½ minutes; Distance =?

v = 108 km/hr = 108 x1000

3600

v= 30m/s

t= Â½ 60 = 30secs

Speed = distance

time

Distance = speed x time

s = 30 x 30

s = 900 m

## CLASS ACTIVITY

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

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

(3)Â Â  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 ACTIVITY

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 thenÂ Â  brought to rest in 10secs with a uniform acceleration. Determine:

(a) The acceleration of the car

(b) The retardation

(c) The distance covered after 5secs

(d) The total distance covered.

Solution

(m/s)

AÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Â Â Â Â Â  B

30

OÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â  E5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  Â Â  20 DÂ Â Â Â Â Â Â  30C

(a)Â Â Â Â  Acceleration =Â Â  AEÂ Â Â  =Â  30

EOÂ Â Â  Â  5

=Â  6m/s2

(b)Â Â Â Â  Retardation = CB Â = 0-30

DCÂ Â Â Â Â  10

= -3m/s2

(c)Â Â Â Â  The distance covered after 5secs = the area is given by area of the triangle

s = Â½ b h

s = Â½ (5) 30

s = 75m

(d)Â Â Â Â  The total distance covered = area of the trapezium OABC

s = Â½ (AB + OC) x h

s = Â½ (15 + 30) x 30

s = 45 x 15

s = 675 m

## MOTION UNDER GRAVITY

AÂ  bodyÂ  movingÂ  with a uniform accelerationÂ  in space does soÂ  underÂ  the influenceÂ  of gravityÂ  with a constantÂ  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 heights 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 + Â½ gt2

s = Â½ gt2 (u=0; initial velocity of an object dropping from a height)

t = âˆš [(2s)/g]

The above relationship can alsoÂ Â  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)

Parabola

t(s)

ButÂ  the graph of s against t2Â  will give usÂ  a straight line through theÂ Â  originÂ  with slopeÂ  Â½ gÂ  from whichÂ  gÂ  canÂ  beÂ  computed

S(m)

X

SlopeÂ  = Â½ g

OÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  t2 (s2)

CALCULATIONS

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

SolutionÂ

v2 = u2 + 2as

u = u; a = -g; v = 0

02 = u2 + 2(-g) s

2gs = u2

s = u2 /2g

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

(i) the time it takes to fall

(ii) the velocityÂ  with which it hitsÂ  the ground

a= +g

u=0

s =20m

t =?

t = âˆš2s/g

t = âˆš (2 x20 /10)

t = 2secs

v = u + gt

v= gt

v = 10 x2

v = 20 m/s

CLASSWORK

1. Define these parameters (a) acceleration (b) velocity (c) displacement
2. A lorry starts from rest and accelerates uniformly until it reaches a velocity of 50 m/s after 10secs. It travels with uniform velocity for 15secs and is brought to rest I 5secs with a uniform retardation. Calculate:
3. a)Â  The acceleration of the lorry
4. b)Â  The retardation
5. c)Â Â  The total distance covered
6. d)Â  The average speed of the lorry

ASSIGNMENT

SECTION A

1. A body is uniformly retarded comes to rest in 10s after travelling a distance of 20m. Calculate its initial velocity (a) 4.0m/s (b) 2.0m/s (c) 20.0m/s (d) 0.5m/s
2. A body accelerates uniformly rest at the rate of 3m/s for 8 seconds. Calculate the distance it covers. (a) 24m (b) 48m (c) 72m (d) 96m.
3. A particle accelerates uniformly from rest at 6m/s2 for 8secs and then decelerates uniformly to rest in the next 5 seconds. Determine magnitude of the deceleration (a) 9.6 m/s2 (b) -9.6 m/s2 (c) 6.9 m/s2 (d) â€“ 6.9 m/s2
4. A car took off from rest and covered a distance of 80m on a straight road in 10s. Calculate the magnitude of its acceleration (a) 1.25m/s2 (b) 1.60 m/s2 (c) 4.00 m/s2 (d) 8.0 m/s2
5. 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.

SECTION B

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