Speed, Velocity & Acceleration

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 

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

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

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

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 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/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.