## LINEAR MOTION FORM THREE

LINEAR MOTION FORM THREE:*Introduction *

Study of motion is divided into two;

**Kinematics****Dynamics**

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

*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).**

LINEAR MOTION FORM THREE

*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 *

*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

*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

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

Time=distance/speed

3000*1000/100

=30,000 seconds.

LINEAR MOTION FORM THREE

*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 *

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

*Calculate, *

*His average speed**His average velocity**His change in velocity for the whole journey*

*Solution *

- Average speed: total distance travelled/total time taken

=800+400/100+80

=1200/180

=6.67m/s

- Average velocity: total displacement/total time

=800-400/180

=400/180

=2.22 m/s due North

- Change in velocity=final-initial velocity

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

=8-5

=3m/s due North

*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

LINEAR MOTION FORM THREE

**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/s ^{2}**

*Examples *

*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/s^{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/s^{2}

Hence retardation is 2.5 m/s^{2 }

**Motion graphs **

Distance-time graphs* *

*Area under velocity-time graph *

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

Distance travelled= average velocity*t

= (0+v/2)*t

=1/2vt

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; *

*Calculate the total distance moved by the car ii.**Find the acceleration of the car at each stage.*

*Solution *

- 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.

- Acceleration= gradient of the graph

Stage A gradient= 20-0/ 10-0 = 2 m/s^{2}

Stage b gradient= 20-20/15-10 =0 m/s^{2}

Stage c gradient= 0-20/21-15 =-3.33 m/s^{2}

**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** 50Hz** i.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.**

LINEAR MOTION FORM THREE

*Examples *

*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

*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/s^{2}

**Equations of linear motion**

The following equations are applied for uniformly accelerated motion; * v = u + at s = ut + ½ at ^{2} v^{2}= u^{2} +2as *

*Examples *

*A body moving with uniform acceleration of 10 m/s*^{2}covers a distance of 320 m.

*if its initial velocity was 60 m/s. Calculate its final velocity. *

*Solution *

V^{2} = u^{2} +2as

= (60) +2×10×320

=3600+6400

= 10,000

Therefore v= (10,000)^{1/2}

v= 100m/s

*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.

*A body is uniformly accelerated from rest to a final velocity of 100m/s in 10 seconds. Calculate the distance covered.**Solution*s=ut + ½ at^{2}

=0 × 10 + ½ ×10 × 10^{2}

= 1000/2=500m

**Motion under gravity.**

LINEAR MOTION FORM THREE

- Free fall

The equations used for constant acceleration can be used to become, *v =u + g t s =ut + ½ gt ^{2} v^{2}= u + 2gs *

- Vertical projection

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

*v =u- g t ……………1 *

*s =ut – ½ gt ^{2} ……2 v^{2}= 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 time taken to fall back to its point of projection.

Using eq. 2 then, displacement =0

**0= ut – ½ gt ^{2} 0=2ut-gt^{2} t (2u-gt)=0 **

Hence**, t=0 or t= 2u/g** t=o corresponds to the start of projection ** t=2u/g** corresponds to the time of flight

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

*Maximum height reached. *

Using equation 3 maximum height, **H _{max}** is attained when v=0 (final velocity). Hence

**v**therefore

^{2}= u^{2}-2gs;- 0=u^{2}-2gH_{max},**2gH _{max}=u^{2} **

* H _{max}=u^{2}/2g *

*Velocity to return to point of projection. *

At the instance of returning to the original point, total displacement equals to zero.

**v ^{2} =u^{2}-2gs** hence

**v**

^{2}= u^{2}Therefore *v=u or v=±u*

*Example *

*A stone is projected vertically upwards with a velocity of 30m/s from the ground. *

*Calculate, *

*The time it takes to attain maximum height**The time of flight**The maximum height reached**The velocity with which it lands on the ground. (take g=10m/s)*

*Solution *

- Time taken to attain maximum height

T=u/g=30/10=3 seconds

- The time of flight

T=2t= 2×3=6 seconds

Or T=2u/g=2×30/10=6 seconds.

- Maximum height reached

H_{max}= u^{2}/2g= 30×30/2×10= 45m

- Velocity of landing (return) v
^{2}= u^{2}-2gs, but s=0,

Hence v^{2}=u^{2}

Therefore v = (30×30)^{1/2}=30m/s

LINEAR MOTION FORM THREE

- Horizontal projection

*The path followed by a body (projectile) is called trajectory. The maximum horizontal distance covered by the projectile is called range. *

The horizontal displacement ‘R’ at a time‘t’ is given by *s=ut+1/2at ^{2}*

Taking u=u and a=0 hence ** R=u t**, is the horizontal displacement and

**is the vertical displacement.**

*h=1/2gt*^{2}**NOTE **

The time of flight is the same as the time of free fall.

*Example *

*A ball is thrown from the top of a cliff 20m high with a horizontal velocity of 10m/s. *

*Calculate, *

*The time taken by the ball to strike the ground**The distance from the foot of the cliff to where the ball strikes the ground.**The vertical velocity at the time it strikes the ground. (take g=10m/s)*

*Solution *

- h= ½ gt
^{2}

20= ½ × 10 × t^{2} 40=10t^{2} t^{2}=40/10=4 t=2 seconds

- R=u t

=10×2

=20m

- v=u +a t=g t

= 2×10=20m/s

LINEAR MOTION FORM THREE

ALL PHYSICS NOTES FORM 1-4 WITH TOPICAL QUESTIONS & ANSWERS

PRIMARY NOTES, SCHEMES OF WORK AND EXAMINATIONS