**SPEED, VELOCITY AND ACCELERATION**

**Speed**

Speed is the time rate of distance. If a car covers a distance, d in a time interval, t, then,

The S.l. unit of speed is metre per second (ms^{-1})

**(a) Average speed**

When the distances covered with time over a particular journey are not steady, possibly due to traffic condition, then the speed can be said to be a non-uniform speed. Under such situation, the total distance covered over the time period is referred to as the average speed. This can be expressed by the relation:

**(b) Instantaneous speed**

This is the speed of an object at a particular measured distance over a short interval of time.

The slope at point P of the graph below is **instantaneous speed**

**Velocity**

Velocity is the time rate of displacement. It is a vector quantity and has the same S.I. unit with speed. It is expressed by the relation:

In other words, velocity can be defined as speed in a specified direction.

**(a) Uniform Velocity**

The velocity of a body is said to be uniform, if it is displaced equally in equal time no matter how small the time interval may be. The uniform velocity of a body can be conveniently determined from the slope of the displacement—time graph. i.e. a graph of displacement against time taken.

**Acceleration**

Acceleration is measured by the rate at which velocity changes with time, that is, acceleration is the time rate of velocity.

If the velocity of an object decreases with time, it is said to be decelerating or retarding. Hence deceleration can be defined as negative acceleration.

By taking

u = initial velocity

v = final velocity

t = time taken

Acceleration can then be expressed thus

**Uniform acceleration**

A body is said to move at uniform acceleration if the rate of change of velocity with time is constant.

Assuming a car starting from rest travels on a level road and gradually increases its velocity at the end of each second, if the velocity attained is as shown below

Then, the uniform acceleration of the car can be determined from the velocity-time graph of the journey as shown below:

**Example:**

A car accelerates uniformly at 2ms^{-2} for 3 minutes. What is the velocity of the car in ms^{-1}?

**Solution:**

First convert minutes to seconds:

3 minutes = 3 x 60seconds = 180 seconds

Acceleration, a = v/t

Therefore, v = at = 2 x 180 = 360ms^{-1}

**Exercises:**

1. A car moving at 10ms^{-1} is brought to rest in 5s. Calculate its retardation (NECO 2019)

A. 50.0 ms^{-2} B. 15.0 ms^{-2} C. 5.0 ms^{-2} D. 2.0 ms^{-2} E. 0.5 ms^{-2}

2. A motorcycle starting from rest moves with uniform acceleration until it attains a speed of 108km/hr after 15s. Find its acceleration.

(Solutions to Exercises are embedded in the evaluation test questions)

**Instantaneous acceleration**

Instantaneous acceleration is the rate of change of velocity with time at a particular instant, for non-uniform acceleration.

**Acceleration due to free fall**

Occurs when a body falls freely under the influence of force of gravity only i.e. without, surrounding air interference

**Determination of acceleration, retardation and distance from velocity – time graph**

A velocity-time graph is simply a graph of velocity plotted against the time taken (see figure below). The area under the line of the graph measures the distance travelled by the car.

**Example:**

The velocity-time graph below illustrates the motion of a body. Use it to answer the questions below:

Determine

(i) the acceleration

(ii) the value of the retardation

(iii) the distance travelled

(iv) the average speed of the motion

**Solutions**

**Exercise**

A train starts from rest and accelerates uniformly at the rate of 2ms^{-2} until it attains a maximum velocity in 20s. It maintains this velocity for 60s and then retards uniformly to rest after a further 20s.

(a) Draw a graph of the motion

(b) Use the graph to determine

(i) The maximum velocity reached

(ii) The retardation and

(iii) The distance covered.

**Equations of Motion**

If an object travelling with a uniform acceleration, a, increases its velocity from a value ‘u’ to a final value ‘v’, in a time t, then,

The basic equations of motion are

v = u + at

s =

s = ut + ½ at^{2}

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

This is negative only when the object undergoes uniform deceleration. Under this condition, the equations becomes

v = u – at

s =

s = ut – ½ at^{2}

v^{2} = u^{2} – 2as

**Exercise**

A car has a uniform velocity of 108km/hr. How far does it travel in ½ minute?

**Solution**

Convert 108 km/hr to m/s

108km/hr = __108 x 1000
__ 3600

= 30 m/s

Since the velocity is uniform, (v=u), therefore we use the 2^{nd} equation of motion

s =

i.e. s =

i.e. s = vt; v = 30ms^{-1}, t = ½ minutes = 30s

s = 30 x 30 = 900m = 0.9km

**Exercise**

The diagram below shows the velocity-time graph of the motion of a body. Calculate the total distance covered. (NECO 2018)

A. 24.0 m B. 36.0 m C. 54.0 m D. 72.0 m E. 108.0 m

**Motion Under gravity
**The motion of objects in space is generally influenced by the action of the earth’s gravity and air resistance.

The equations of motion under gravity are:

(1) v = u +gt

(2) h = ut + ½ gt^{2}

(3) v^{2} = u^{2} + 2gh

where h = height and g = acceleration due to free fall. (This is approximately equal to 10ms^{-2}).

NB: Negative **g (acceleration due to free fall)** is applied when the objects motion is directed upwards. Then the equations become:

(1) v = u – gt

(2) h = ut – ½ gt^{2}

(3) v^{2} = u^{2} – 2gh

**Acceleration Of Free Fall Due To Gravity**

Objects that fall freely have accelerated motion because of the action of the earth’s gravity on them.

For example: if a piece of stone is held, and let go from the top of a high-rise building, it tends to fall from that height more and more faster until it hits the ground. It has also been observed that objects no matter their difference in mass fall under gravity with the same acceleration and at the same time, if the air resistance is neglected.

**Vertical Projection**

If a piece of stone or ball is projected vertically upwards, its projection velocity tends to decrease more slowly until it comes momentarily to a stop. It then falls back faster and faster till it hits the plane of projection.

The decrease in velocity as the ball rises higher up, is due to the effect of gravity. However, the maximum height an object under this condition can attain depends on the magnitude of the initial upward velocity used in projecting it.

if h_{max} = maximum height attained

u = Initial upward velocity

g = acceleration of free fall

Then,

Time t, taken to attain the maximum height would be obtained from

v = u – gt …..equation 1

**Note** that at maximum height, the object usually comes to a momentary stop. **Note** also that g is negative (-) because the object’s motion is directed upwards

Hence,

final velocity, v = 0

therefore from equation 1 above,

0 = u – gt

i.e. gt = u

therefore, t = ………eqn 2

**To determine the time of flight, T**

Note that the time of flight is the time required for the object to be in air or in flight. This time, T is twice the time required for the object to reach the maximum height.

Hence,

T = 2t ……………eqn 3

substituting the value of t from eqn 2 into eqn 3

T = 2( )

T = ……………eqn 4

**To determine the maximum height, attained**

from eqn of motion under gravity,

v^{2} = u^{2} – 2gh_{max}

at maximum height, v = 0

0 =u^{2} – 2gh_{max},

2gh_{max} = u^{2}

therefore, h_{max} =

**Exercise:**

A ball is projected vertically upward from the top of a tower 80m high with a velocity of 20ms^{-1}. Calculate the height reached by the ball above the ground. Take g = 10ms^{-2} (NECO 2018)

A. 20m B. 60m C. 81m D. 82m E. 100m

**Exercise**

A tennis ball is thrown vertically upwards from the ground with a velocity of 50m/s.

Calculate:

(i) the maximum height reached

(ii) the time to reach maximum height

(iii) the time of flight

**Horizontal Projection**

When an object is projected horizontally with a horizontal velocity, u, from a height, h, it will

(i) describe a path that is half a parabola as shown in the figure below

(ii) take the same time to reach the ground as though it is dropped vertically from the same height.

**To determine the height h, from which an object is projected.**

Note that the projection velocity is horizontal, as such, velocity in the vertical direction = 0 By applying the equation

**Example**:

A ball is projected horizontally from the top of a hill with a velocity of 20ms^{-1}. If it reaches the ground 4 seconds later, what is the height of the hill?

**Solution**

The height of the hill = ½ gt^{2}

= ½ x 10 x 4^{2}

= 80m

**Exercise**

A stone is projected horizontally from the top of a tower with a speed of 5ms^{-1}. It lands on the ground at a distance of 20m from the foot of a tower. Calculate the height of the tower above the ground (Take g = 10ms^{-2})

A. 40m B. 80m C. 100m D. 160m

**Resultant velocity of a projectile**

When a projectile is in flight, its resultant velocity, v which is tangential to the direction of flight is made up of

(i) horizontal component velocity, V_{x} = horizontal velocity of projection, U

(ii) vertical component velocity, V_{y} = vertically downward velocity, (gt).

(since u = 0 in the vertical direction)

u = horizontal velocity of projection.

**Motion Down an Inclined Plane**

If the friction between an object and the inclined plane in which it is in contact with is neglected, the object would tend to slip down the inclined surface faster because of the component effect of acceleration due to gravity.

**Scroll Down to Select Page 7 for the next topic – ELECTRIC CHARGES**

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