2nd Equation of motion.

Displacement: β€œS or x” is length moved in specified direction
From the definition of Displacement. π·π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘=(π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦)Γ—(π‘‡π‘–π‘šπ‘’)

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This is called the second equation of motion. This equation is mainly used when the question involves distance and time.

Example: 1

A body starts from rest and accelerates uniformly at 2ms-2 for 3s. Calculate the total distance travelled.

solution

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Calculations involving deceleration or Retardation
When calculating a problem involving deceleration; it should be remembered that the value of β€œa” should be negative.

Example 2:
A body moving at 40m/s decelerates uniformly for 20s at 3m/s2. Calculate distance covered.

Solutions

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Example 3:
A car traveling at 40m/s is uniformly decelerated to 25m/s for 5s. Calculate the total distance covered.

Solution

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Third Equation of motion
From: π·π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘=(π΄π‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦)Γ—(π‘‡π‘–π‘šπ‘’).
Making β€˜t’ the subject of the formula in the first equation of motion and substituting it in here, we get;

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This is called the third equation of motion
This equation is applied when time is not given and not required.

Example 1:
Calculate the final (maximum) velocity of a body traveling at 4m/s. When it accelerates at 2m/s2 and covers a distance of 5m.

Solution

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Example 2:
A body traveling at 90km/hr is retarded to rest at 20m/s2. Calculate the distance covered.

Solution

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Graphical presentation of uniform velocity and uniform acceleration.

Uniform velocity can be represented on a 2 type of graphs.
i) Velocity against time graph
ii) Distance against time.

i) velocity against time graph

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Note:
When a body maintains the same speed, it implies that it moves with uniform velocity.

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OA- uniform acceleration
AB- uniform velocity
BC- uniform acceleration
CD- uniform deceleration or uniform retardation

The slope of a velocity time graph gives the acceleration of the body. ie:

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The Area under any stage or section of velocity time graph gives the distance covered during that time.

Drawing a velocity against time graph

This involves the following steps:

Divide the motion into stages basing on the timing.

Obtain the initial velocity (u) and final velocity (v) for each stage.

The final velocity for one stage becomes the initial for the next stage.

Example 1:
A cyclist starts from rest and accelerate uniformly at 1m/s2 for 20s. Then he maintains the maximum speed so reached for 1 minute and finally decelerates to rest uniformly for 10s.
i) Draw a velocity against time graph for the body.
ii) Calculate the total distance travelled.

solutions

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(i) A velocity against time graph for the motion.

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ii) Total distance travelled

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Example 2:
A car traveling at 10m/s is uniformly accelerated for 4s at 2m/s2. It then moves with a constant speed for 5s after which it is uniformly brought to rest in another 3s.
i) Draw a velocity against time graph.
ii) Calculate the total distance travelled.

solutions

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Then from;

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i) Total distance travelled

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Note: Distance covered during stage A can also be obtained by dividing the area A into a triangle and a rectangle and then finding the sum of the two areas.

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Example 3:
A body moving with uniform velocity:
A car travels at a velocity of 20m/s for 6s. It is then uniformly brought to rest in 4s.
i) Draw a velocity against time graph.
ii) Calculate the retardation
iii) Find the total distance traveled
iv) Calculate the average speed of the body

Solution

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ii) The retardation
Retardation or deceleration occurs in region B.

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iii) Total distance travelled

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Motion under gravity (Falling bodies)

In a vacuum, all bodies fall at the same rate. However, in atmosphere different bodies fall at different rate because the air resistance is greater to light objects.

Acceleration due to gravity, g.
Acceleration due to gravity is the change in velocity with time for body falling freely under the force of gravity.

Note: Acceleration due to gravity varies from place to place because:
The earth is not a perfect sphere
The earth is always rotating

All bodies thrown upwards or falling freely in the earth’s surface, have a constant acceleration called Acceleration due to gravity, 𝑖.𝑒:𝒂=π’ˆ=𝟏𝟎 π’Žπ’”βˆ’πŸ.

Since the gravitational force acts vertically down wards, ie accelerates all objects down wards towards the earth’s surface. Thus for downward motion (falling objects), 𝒂=+π’ˆ=+𝟏𝟎 π’Žπ’”βˆ’πŸ. And for upward motion (objects thrown upwards), 𝒂=βˆ’π’ˆ=βˆ’πŸπŸŽ π’Žπ’”βˆ’πŸ.

Projectile motion
A projectile is a particle which has both vertical and horizontal motions when thrown in air.
Consider a body thrown vertically upwards from A.

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In projectiles, the horizontal and vertical motions are handled separately but simultaneously. The horizontal velocity of the body in motion remains the same throughout since there is no acceleration due to gravity in the horizontal.

First equation of motion

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Third equation of motion
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Maximum Height, π’”π’š=π‡π¦πšπ±: Is the highest vertical distance attained by a projectile.
At Hmax , 𝑣𝑦=0:

Time of Flight, T: Is the total time taken for a projectile to move from origin until it lands. This time is twice the time taken to reach the maximum height.
If t is the time taken to reach the maximum height, then

T=2t

Distance – Time or Displacement –Time Graphs for a body thrown vertically upward.

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Distance- Time or Displacement –Time Graph for a body falling freely from rest.
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Displacement –Time Graph for a body thrown vertically upwards from appoint above the ground.

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Speed- Time and Velocity –Time Graphs For a body thrown vertically upwards.

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The speed of the object decreases as it goes higher. At maximum height reached the speed is zero because the object is momentarily at rest and when the object starts to fall the speed increases.

The velocity decreases upwards and it is zero at the maximum height. However the velocity increases downwards and negative because of the change in direction.

Experiment to measure acceleration due to gravity.

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A tape is passed through a ticker-timer and attached to a 10g- mass.

The ticker-timer makes dots on the tape at an interval determined by the frequency of the mains supply. i.e 𝑇=1𝑓. This is the time taken to make one space (2 dots).

The distance S between the first dot to the last dot made just before the mass hits the ground is measured using a metre-rule.

The time,t taken to make n- spaces in distance S is calculated from: 𝑑=𝑛𝑇.

The acceleration due to gravity, g is then calculated from

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Experiment to measure acceleration due to gravity

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A pendulum bob is suspended from a clamp using an inextensible string as shown in the diagram above.

The length of the string of the pendulum bob β€˜l’ is adjusted such that l=0.3m.

The bob is the slightly displaced through a small angle and released

The stop clock is started and the time taken to make 20 oscillations (20T) is measured and recorded

The period time T for a single oscillation is calculated and recorded.

The experiment is repeated for other increasing values of l, and the corresponding values of 20T, T and T2 calculated and tabulated.
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A graph of T2 against l is plotted. It is a straight line graph through the origin and its slope, S is calculated.

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NOTE: Experiments have shown that the periodic time T does not depend on the mass of the bob, but it depends on the length of pendulum l bob and acceleration due to gravity g at that point. i.e:

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Example1:
A stone falls from rest from the top of a high tower. Calculate the velocity after 2s.

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Example 2:
An object is dropped from a helicopter. If the object hits the ground after 2s, calculate the height from which the object was dropped.

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Example 3:
An object is dropped from a helicopter at a height of 45m above the ground.
a) If the helicopter is at rest, how long does the object take to reach the ground and what is its velocity on arrival?

Solution
u= 0 ms-1
a = g =10 ms-2
t = ?
s = 45m

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b) If the helicopter had a velocity of 1ms-1 when the object was released, what would be the final velocity of the object?

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Example 4:
An object is released from an aircraft traveling horizontally with a constant velocity of 200ms-1 at a height of 500m.
a) Ignoring air resistance, how long it takes the object to reach the ground?

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b) Find the horizontal distance covered by the object between leaving the aircraft and reaching the ground.

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Note: For a body thrown vertically up ward, the time taken to reach the maximum vertical height is equal to the time taken for the body to fall from maximum height.

Note; If a body is not falling freely but there is air resistance R then the acceleration of the body can be calculated from:-
ma = mg – R , where m is the mass of the body

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Example 4:
An object of 2kg is dropped from a helicopter at a height 45m above the ground. If the air resistance is 0.8N, calculate the:
i) acceleration of the body
ii) velocity with which the body hits the ground

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