**Physics Notes Form 3**

Physics Notes Form 3

Physics Form Three

Chapter One

Linear Motion

**Introduction**

**Study of motion is divided into two;**

1. Kinematics

2. Dynamics

In kinematics forces causing motion are disregarded while dynamics deals with motion of objects and the forces causing them.

**I. Displacement**

Distance moved by a body in a specified direction is called displacement. It is denoted by letter‘s’ and has both magnitude and direction. Distance is the movement from one point to another. The Si unit for displacement is the metre (m).

**II. Speed**

This is the distance covered per unit time.

Speed= distance covered/ time taken. Distance is a scalar quantity since it has magnitude only.

The SI unit for speed is metres per second(m/s or ms^{-1})

**Average speed**= total distance covered/total time taken Other units for speed used are Km/h.

**Examples**

1. A body covers a distance of 10m in 4 seconds. It rests for 10 seconds and finally covers a distance of 90m in 60 seconds. Calculate the average speed.

**Solution**

**Total distance covered** =10+90= 100m

**Total time taken** =4+10+6= 20 seconds

**Therefore average speed** = 100/20= 5m/s

2. Calculate the distance in metres covered by a body moving with a uniform speed of 180 km/h in 30 seconds.

**Solution**

**Distance covered**=speed*time

=180*1000/60*60=50m/s

=50*30

=1,500m

3. Calculate the time in seconds taken a by body moving with a uniform speed of 360km/h to cover a distance of 3,000 km?

**Solution**

Speed:360km/h=360*1000/60*60=100m/s

**Time**=distance/speed

3000*1000/100

=30,000 seconds.

**III. Velocity**

This is the change of displacement per unit time. It is a vector quantity.

**Velocity**=change in displacement/total time taken

The SI units for velocity are m/s

**Examples**

1. A man runs 800m due North in 100 seconds, followed by 400m due South in 80 seconds. Calculate,

a. His average speed

b. His average velocity

c. His change in velocity for the whole journey

**Solution**

**a. Average speed**: total distance travelled/total time taken

=800+400/100+80

=1200/180

=6.67m/s

**b. Average velocity**: total displacement/total time

=800-400/180

=400/180

=2.22 m/s due North

**c. Change in velocity**=final-initial velocity

= (800/100)-(400-80)

=8-5

=3m/s due North

2. A tennis ball hits a vertical wall at a velocity of 10m/s and bounces off at the same velocity. Determine the change in velocity.

**Solution**

Initial velocity(u)=-10m/s

Final velocity (v) = 10m/s

Therefore change in velocity= v-u

=10- (-10)

=20m/s

**IV. Acceleration**

This is the change of velocity per unit time. It is a vector quantity symbolized by ‘a’. Acceleration ‘a’=change in velocity/time taken= v-u/t The SI units for acceleration are m/s2 Examples 1. The velocity of a body increases from 72 km/h to 144 km/h in 10 seconds.

Calculate its acceleration.

**Solution**

Initial velocity= 72 km/h=20m/s

Final velocity= 144 km/h=40m/s

Therefore ‘a’ =v-u/t

= 40-20/10

2m/s2

2. A car is brought to rest from 180km/h in 20 seconds. What is its retardation?

**Solution**

Initial velocity=180km/h=50m/s

Final velocity= 0 m/s

A = v-u/t=0-50/20

= -2.5 m/s2

Hence retardation is 2.5 m/s2

**Motion graphs**

Distance-time graphs

a)

b)

**Area under velocity-time graph**

Consider a body with uniform or constant acceleration for time‘t’ seconds;

This is equivalent to the area under the graph. The area under velocity-time graph gives the distance covered by the body under‘t’ seconds.

**Example**

A car starts from rest and attains a velocity of 72km/h in 10 seconds.

It travels at this velocity for 5 seconds and then decelerates to stop after another 6 seconds.

Draw a velocity-time graph for this motion. From the graph;

i. Calculate the total distance moved by the car

ii. Find the acceleration of the car at each stage.

**Solution**

a. From the graph, total distance covered= area of (A+B+C)

=(1/2×10×20)+(1/2×6×20)+(5×20)

=100+60+100

=260m

Also the area of the trapezium gives the same result.

b. Acceleration= gradient of the graph

Stage A gradient= 20-0/ 10-0 = 2 m/s2

Stage b gradient= 20-20/15-10 =0 m/s2

Stage c gradient= 0-20/21-15 =-3.33 m/s2

**Using a ticker-timer to measure speed, velocity and acceleration.** It will be noted that the dots pulled at different velocities will be as follows;

Most ticker-timers operate at a frequency of 50Hzi.e. 50 cycles per second hence they make 50 dots per second. Time interval between two consecutive dots is given as,

**1/50 seconds= 0.02 seconds. This time is called a tick.**

The distance is measured in ten-tick intervals hence time becomes 10×0.02= 0.2 seconds.

**Examples**

a. A tape is pulled steadily through a ticker-timer of frequency 50 Hz.

Given the outcome below, calculate the velocity with which the tape is pulled.

**Solution**

Distance between two consecutive dots= 5cm

Frequency of the ticker-timer=50Hz

Time taken between two consecutive dots=1/50=0.02 seconds

Therefore, velocity of tape=5/0.02= 250 cm/s

b. The tape below was produced by a ticker-timer with a frequency of 100Hz. Find the acceleration of the object which was pulling the tape.

**Solution**

Time between successive dots=1/100=0.01 seconds

Initial velocity (u) 0.5/0.01 50 cm/s

Final velocity (v) 2.5/0.01= 250 cm/s

Time taken= 4 ×0.01 = 0.04 seconds

Therefore, acceleration= v-u/t= 250-50/0.04=5,000 cm/s2

**Equations of linear motion**

The following equations are applied for uniformly accelerated motion;

v = u + at

s = ut + ½ at2

v2= u2 +2as

**Examples**

1. A body moving with uniform acceleration of 10 m/s2 covers a distance of 320 m. if its initial velocity was 60 m/s. Calculate its final velocity.

**Solution**

V2 = u2 +2as

= (60) +2×10×320

=3600+6400

= 10,000

Therefore v= (10,000)1/2

v= 100m/s

2. A body whose initial velocity is 30 m/s moves with a constant retardation of 3m/s. Calculate the time taken for the body to come to rest.

**Solution**

v = u+at

0= 30-3t

30=3t

t= 30 seconds.

3. A body is uniformly accelerated from rest to a final velocityof 100m/s in 10 seconds. Calculate the distance covered.

**Solution**

s=ut+ ½ at2

=0×10+ ½ ×10×102

= 1000/2=500m

**Motion under gravity.**

**1. Free fall**

The equations used for constant acceleration can be used to become,

v =u+gt

s =ut + ½ gt^{2}

v2= u+2gs

**2. Vertical projection**

Since the body goes against force of gravity then the following equations hold

v =u-gt ……………1

s =ut- ½ gt2 ……2

v2= u-2gs …………3

N.B time taken to reach maximum height is given by the following

t=u/g since v=0 (using equation 1)

**Time of flight**

The time taken by the projectile is the timetaken to fall back to its point ofprojection. Using eq. 2 then, displacement =0

0= ut- ½ gt^{2}

0=2ut-gt2

t(2u-gt)=0

Hence, t=0 or t= 2u/g

t=o corresponds to the start of projection

t=2u/gcorresponds to the time of flight

The time of flight is twice the time taken to attain maximum height.

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