### 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} âœ…

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

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

= 20ms^{-1} âœ…

**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 2Â½ 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 =Â \( \frac {change\: in\: displacement}{time} \)

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

= 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/s^{2}

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 is zero in some cases.

**Also Note:** When a body starts from rest, its initial velocity, u, is 0, When a body comes to rest (stops moving) its 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/s^{2}

**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/s^{2}

**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/sÂ², find the final velocity of the car.

Acceleration = 0.25m/sÂ²

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

this is so self explanatory

Highly helpful…

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Great lesson