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SS1: PHYSICS – 1ST TERM

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  1. Introduction to Physics | Week 1
    4Topics
    |
    1 Quiz
  2. Measurement | Week 2
    3Topics
  3. Measurement of Mass | Week 3
    6Topics
    |
    1 Quiz
  4. Motion | Week 4
    5Topics
    |
    1 Quiz
  5. Velocity-Time Graph | Week 5
    4Topics
    |
    1 Quiz
  6. Causes of Motion | Week 6
    5Topics
    |
    1 Quiz
  7. Work, Energy & Power | Week 7
    3Topics
  8. Energy Transformation / Power | Week 8
    3Topics
    |
    1 Quiz
  9. Heat Energy | Week 9
    5Topics
    |
    1 Quiz
  10. Linear Expansion | Week 10
    6Topics
    |
    1 Quiz
Lesson 4, Topic 5
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Speed, Velocity & Acceleration

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Speed:

Speed is the distance covered per time taken or the rate of change of distance. It is a scalar quantity and measured in km per hour or metre per second (m/s or \( \scriptsize ms^{-1}\)).

Speed =  \( \frac {distance}{time} \)

Average Speed =   \(\frac{Total \: distance \:covered}{total\:time\: taken} \)

Uniform Speed: A body is said to move with uniform speed if the time rate of change in distance is constant.

Speed in  \( \frac {km}{hr}\)or km/hr can easily be converted to m/s or ms-1.

This is achieved as follows:

Speed =   \( \frac {distance \: in \: km}{time\: in\: hour} = \frac {1km}{1hr} \)

But 1km = 1000m, 1hr = 3600s

Therefore, To convert km/hr to m/s, the conversion factor is 

\( \frac {1000}{3600} \scriptsize \: \times \: (Speed \: in \: km/h)\; = \: Speed\: in\: ms^{-1} \)

Example 1

i) Convert 36 \( \frac {km}{hr} \)  to ms-1

Solution:   \( \frac {1000}{3600}\scriptsize \times 36 \)

= 10ms-1

ii) Convert 72 \( \frac {km}{hr} \)  to ms-1

Solution:   \( \frac {1000}{3600} \scriptsize \times 72 \)

= 20ms-1

Note: To convert from m/s to km/h multiply by \( \frac{3600}{1000} \)

20ms-1 to km/hr

= \( \scriptsize 20 \: \times \: \normalsize \frac{3600}{1000} \)

= 20 x 3.6

= 72km/hr

Example 2

A train covers a distance of 100km in half an hour. What is the average speed of the train in

a. km/hr

b. m/s

Solution:

a. time = \( \frac{1}{2} \scriptsize hour \\ \scriptsize = 0.5 hour \)

Average Speed = \( \frac {distance \: in \: km}{time\: in\: hour} \)

= \( \frac{100km}{0.5hr} \\ \scriptsize = 200km/hr\)

b. Speed = \( \frac {distance \: in \: km}{time\: in\: hour} \\ = \frac {1km}{1hr} \\ = \frac{1000m}{3600s} \)

1km = 1000m, 1hr = 3600s

200km/hr = \( \scriptsize 200 \: \times \: \normalsize \frac{1000}{3600} \)

= 55.56 m/s

Example 3

Calculate the distance traveled by a train in 2½ minutes if the speed of the train is 2.25 m/s.

i. time = \( \scriptsize 2 \frac{1}{2} hour \\ \scriptsize = 2 \frac{1}{2} \; \times \; 60 \\ \frac{5}{2} \scriptsize \; \times \; 60 \\ = \scriptsize 30 \; \times \; 50 \)

= 150 seconds

Speed = \( \frac {distance }{time} \)

Therefore distance = \( \scriptsize Speed \: \times \: Time \)

= 2.25 x 150

= 337.5 metres

Velocity:

Velocity is defined as the rate of change of displacement with respect to time. It is a vector quantity and its unit is metre per second, m/s or \( \scriptsize ms^{-1} \)

V =  \( \frac {change\: in\: displacement}{time} \)

=\( \frac {metre}{second}\)

= metre per second (m/s)   = ms-1

u – Initial velocity of a body before moving

v – Final velocity of a body

Average Velocity is the sum of initial and final velocity divided by 2

Average Velocity =  \( \frac {initial \:+\: final\;velocity}{2} \)

= \( \frac {u\: + \: v}{2} \)

Uniform Velocity: A body is said to move with uniform velocity if the time rate of change in displacement is constant.

Example 4

Calculate the velocity of a train whose displacement in 3 minutes is 500 meters.

Displacement = 500 metres

Time = 3mins

We first have to convert the time from mins to seconds = 3 x 60 = 180 seconds

V =  \( \frac {change\: in\: displacement}{time} \)

Substitute the Values into the equation

V =  \( \frac {500}{180} \)

Velocity = 2.78 m/s

Acceleration:

Acceleration is the time rate of change in velocity with time. It is a vector quantity and is measured in ms-2 or m/s2

Acceleration, a, =   \( \frac {increase\: in\: velocity}{time}\) 

a =   \( \frac {v \: – \: u}{t} \)

Where;

a = acceleration

v = Final Velocity

u = Initial Velocity

t = time taken

Retardation: This is the time rate of decrease in Velocity. 

Retardation:   =  \( \frac {decrease \: in \: velocity}{time} = \frac {v \: – \: u}{t} \) 

Note:

Retardation means a reduction in velocity. In retardation, the initial velocity is always larger than the final velocity which in some cases is zero.

Also Note:

When a body starts from rest, it’s initial velocity, u, is 0, When a body comes to rest (stops moving) it’s final velocity, v, is 0

Uniform Acceleration: A body is said to experience uniform acceleration if its time rate of increase in velocity is constant.

Example 5

a. A train travels at a speed of 7.2 m/s for 200 seconds. Calculate the acceleration of the train.

Speed = 7.2m/s, Time = 200 seconds

acceleration, a, = \( \frac {velocity}{time}\) 

a = \( \frac {7.2}{200}\) 

a= 0.036 m/s2

b. A Ferrari sports car starts from rest and is moves to a velocity of 30m/s in 3 minutes. Find the acceleration of the car.

The car accelerates from rest so the initial velocity, u, = 0 m/s

Final velocity v = 30 m/s

Time = 3mins = 3 x 60 = 180 seconds

Acceleration, a, =   \( \frac {increase\: in\: velocity}{time}\) 

a =   \( \frac {v \: – \: u}{t} \)

Substitute in the values from the question

a =   \( \frac {30 \: – \: 0}{180} \)

a =   \( \frac {30}{180} \)

a = 0.167 m/s2

c. A car accelerated from 8.0 m/s to a velocity of y m/s in 2 minutes. If the acceleration of the car was 0.25 m/s², find the final velocity of the car.

Acceleration = 0.25m/s²

Initial Velocity, u = 8.0 m/s

Time taken = 2 mins

= 2 x 60

= 120 seconds

Final Velocity, v = ?

Acceleration, a, =   \( \frac {increase\: in\: velocity}{time}\) 

a =   \( \frac {v \: – \: U}{t} \)

make v the subject of the formula

multiply both sides by t

at = v – u

Add u to both sides

at + u = v – u + u

:- at + u = v

:- v = at + u

Substitute in the values

v = 0.25 x 120 + 8

v = 30 + 8 = 38 m/s

d. A train decelerates from 100 km/h to 40 km/h in 2 minutes. Determine the retardation of the train.

Initial Velocity u = \( \frac{100km}{h} \)

convert to m/s

= \( \frac{1000}{3600}\scriptsize \: \times \: 100 \)

= \( \scriptsize 22.78 m/s \)

Final Velocity v = \( \frac{40km}{h}\)

= \( \frac{1000}{3600}\scriptsize \: \times \: 40 \)

= \( \scriptsize 11.11 m/s \)

Time taken = 2 mins

= 2 x 60 = 120 seconds

Retardation = \( \frac {v \: – \: u}{t} \)

Substitute in the values into the formula

= \( \frac {11.11 \: – \: 22.78}{120} \)

= \( \frac {-11.67}{120} \)

Retardation = -0.09725

The negative sign shows that it is decelerating.

Evaluation Questions

1. Define motion

2. List the types of motion you know

3. Distinguish between circular motion and oscillatory motion with one example each

4. Convert the following to ms-1

(i) 36km/h (ii) 108km/h (iii) 56km/h (iv) 96km/h

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