14.0.0: LINEAR ACCELERATION
CONCEPT OF ACCELERATION:
Acceleration is the time rate of increase in the velocity of an object. It is a scalar quantity. The s.I. unit of acceleration is meter per second square, m/s².
UNIFORM ACCELERATION:
uniform acceleration is the constant time rate of increase in the velocity of an object no matter how small the time may be.
FORMULA OF ACCELERATION:
Acceleration = change in velocity / time ➡ acceleration = increase in velocity / time
Acceleration = ( V - U ) / t₂ - t₁ ➡. Acceleration = Velocity / time
APPLICATION OF FORMULA:
EXAMPLE 1:
A train travels at the speed of 5.2 m/s for 200 seconds. Calculkate the acceleration of the train.
SOLUTION:
Data given in the question:
Speed = 2.5 m/s, time = 200 seconds.
Formula: acceleration = velocity / time
Substitution : acceleration = 2.5 / 200 = 0.0125 m/s²
EXAMPLE 2:
A car start from rest and accelerates to 13.2 m/s for 2 minutes. Find the acceleration of the car.
SOLUTION:
Data given in the question:
Since the car start from rest to accelerate, initial velocity V₁ = 0 m/s,
final velocity V₂ = 13.2m/s, time = 2 minutes = 2 * 60 seconds = 120 seconds.
Formula : acceleration = change in velocity / time ➡ acceleration =( V - U ) / time
Substitution : acceleration = 13.2 - 0 / 120 ➡ acceleration = 13.2 /120
Acceleration = 0.11 m/s²
EXAMPLE 3:
A car accelerated from 5.0 m/s to Ym/s in 1 minute. If the acceleration of the car was 0.15 m/s², find the final velocity of the car.
SOLUTION:
Data given in the question:
Acceleration = 0.15 m/s², initial velocity U = 5 m/s, final velocity V = Ym/s,
time = 1 minute = 1 * 60 second = 60 seconds.
Formula: acceleration = ( V - U ) / time , ➡
Substitution : 0.15 = ( Y - 5) / 60 ➡ 0.15 * 60 = Y - 5 ➡ Y = 0.15 * 60 + 5
➡ Y = 9 + 5 = 14 m/s. Final velocity Y = 14 m/s
VELOCITY - TIME GRAPH OF ACCELERATION:
Velocity - time graph of acceleration is obtained by plotting different values of velocities on the vertical axis against different values of time on the horizontal axis. This is shown in the figure below:
a
b
Figure a is the velocity- time graph of an object that accelerated from rest while figure b is the velocity - time graph of an object that is moving with uniform velocity with time. In figure b, the velocity of the object is constant, that is the velocity remain the same with time.
SLOPE OF VELOCITY-TIME GRAPH OF ACCELERATION:
In figure a, the slope = 🔼V / 🔼t ➡ slope = change in velocity / change in time
Slope = ( Final velocity - initial velocity ) / time ➡ slope = (V₂ - V₁) / t
Slope = Velocity / time ( which is the formula of acceleration)
Therefore, in figure a, the slope of the graph represent the acceleration of the object.
Therefore, take note that what ever the value that you obtained when calculating the slope is the acceleration of the object.
In figure b, the slope = 🔼V / 🔼t ➡ slope = change in velocity / change in time
Slope = ( Final velocity - initial velocity ) / time ➡ slope = (V₂ - V₁) / t
Slope = Velocity / time ( which is the formula of acceleration) . in figure b, V₂ & V₁ are the same because the velocity of the object is constant throughout the time frame. Therefore, V₂ = V₁. Slope = (V₂ - V₂) / t. ➡ slope = 0 / t. ➡ slope = 0.
Therefore, since slope = 0, acceleration of the object is equal to zero in velocity - time graph of uniform velocity. ( I.e acceleration = 0).
DECELERATION OR RETARDATION:
Retardation is the time rate of decrease in the velocity of an object. Deceleration or retardation is negative acceleration. The s.I.unit of retardation is m/s²
UNIFORM RETARDATION:
Uniform retardation is the constant time rate of decrease in the velocity of an object. The unit of retardation is m/s²
FORMULA FOR CALCULATING RETARDATION:
retardation = 🔼V / 🔼 t ➡ retardation = change in velocity / change in time.
➡ retardation = ( Final velocity - initial velocity ) / time ➡ retardation = (U - V) / t
You must take note that, retardation means reduction in velocity. In retardation, the initial velocity is always larger than the final velocity which in some cases are zero.
APPLICATION OF FORMULA :
EXAMPLE 1:
Attain decelerates from 25 m/s to 15 m/s in 2 minutes. Determine the retardation of the train.
SOLUTION:
Data given in the question:
Initial velocity U = 25 m/s , final velocity V = 15 m/s, time = 2 minutes = 2 * 60 seconds. Time = 120 seconds.
Forumla : retardation = ( V - U ) / time
Substitution: retardation = ( 15 - 25 ) / 120. ➡ retardation = - 10 / 120
➡ retardation = 0.0833 m/s² or acceleration = - 0.083 m/s²
Example 2:
A car which travels at the rate of 35 m/s slow down to a stop when the brakes where applied. If the time taken is 1½ minutes, calculate the deceleration seconds.
SOLUTION:
Data given in the question:
Initial velocity U = 35 m/s, final velocity V = 0 m/s ( because the car slow down and stop ), time = 1½ minutes, time = 1½ * 60 seconds = 3/2 * 60 seconds = 90 seconds
Formula: deceleration = ( V - U ) / time
Substitution: deceleration = ( 0 - 35 ) / 90 = ➡ deceleration = - 35 / 90
➡ deceleration = 0.388 m/s² or acceleration = - 0.388 m/s²
Example 3:
Given that the acceleration of a car is - 0.25 m/s² when the car slow down from its initial velocity Y m/s to 1.25 m/s in 35 seconds, calculate the initial velocity Y of the car.
SOLUTION:
Data given in the question:
Acceleration = - 0.25 m/s², initial velocity U = Y m/s, final velocity V = 1.25 m/s ,
time = 35 seconds.
Formula: deceleration = ( V - U ) / time
Substitution: - 0.25 = ( 1.25 - U ) / 35 ➡ - 0.25 * 35 = 1.25 - U. ➡ - 0.00714 = 1.25 - U
Make U the subject : U = 1.25 + 0.00714. ➡. U = 1.25714 m/s
EXERCISES:
VELOCITY-TIME GRAPH :
The velocity - time graph is obtained by plotting different values of velocity on the vertical axis against different values of time on the horizontal axis.
a. 
b.
The three separate graphs are combined to get the velocity time graph for the combined motion. The shape of the graph is a shape of a trapezium.
CONCEPT OF ACCELERATION:
Acceleration is the time rate of increase in the velocity of an object. It is a scalar quantity. The s.I. unit of acceleration is meter per second square, m/s².
UNIFORM ACCELERATION:
uniform acceleration is the constant time rate of increase in the velocity of an object no matter how small the time may be.
FORMULA OF ACCELERATION:
Acceleration = change in velocity / time ➡ acceleration = increase in velocity / time
Acceleration = ( V - U ) / t₂ - t₁ ➡. Acceleration = Velocity / time
APPLICATION OF FORMULA:
EXAMPLE 1:
A train travels at the speed of 5.2 m/s for 200 seconds. Calculkate the acceleration of the train.
SOLUTION:
Data given in the question:
Speed = 2.5 m/s, time = 200 seconds.
Formula: acceleration = velocity / time
Substitution : acceleration = 2.5 / 200 = 0.0125 m/s²
EXAMPLE 2:
A car start from rest and accelerates to 13.2 m/s for 2 minutes. Find the acceleration of the car.
SOLUTION:
Data given in the question:
Since the car start from rest to accelerate, initial velocity V₁ = 0 m/s,
final velocity V₂ = 13.2m/s, time = 2 minutes = 2 * 60 seconds = 120 seconds.
Formula : acceleration = change in velocity / time ➡ acceleration =( V - U ) / time
Substitution : acceleration = 13.2 - 0 / 120 ➡ acceleration = 13.2 /120
Acceleration = 0.11 m/s²
EXAMPLE 3:
A car accelerated from 5.0 m/s to Ym/s in 1 minute. If the acceleration of the car was 0.15 m/s², find the final velocity of the car.
SOLUTION:
Data given in the question:
Acceleration = 0.15 m/s², initial velocity U = 5 m/s, final velocity V = Ym/s,
time = 1 minute = 1 * 60 second = 60 seconds.
Formula: acceleration = ( V - U ) / time , ➡
Substitution : 0.15 = ( Y - 5) / 60 ➡ 0.15 * 60 = Y - 5 ➡ Y = 0.15 * 60 + 5
➡ Y = 9 + 5 = 14 m/s. Final velocity Y = 14 m/s
VELOCITY - TIME GRAPH OF ACCELERATION:
Velocity - time graph of acceleration is obtained by plotting different values of velocities on the vertical axis against different values of time on the horizontal axis. This is shown in the figure below:
abFigure a is the velocity- time graph of an object that accelerated from rest while figure b is the velocity - time graph of an object that is moving with uniform velocity with time. In figure b, the velocity of the object is constant, that is the velocity remain the same with time.
SLOPE OF VELOCITY-TIME GRAPH OF ACCELERATION:
In figure a, the slope = 🔼V / 🔼t ➡ slope = change in velocity / change in time
Slope = ( Final velocity - initial velocity ) / time ➡ slope = (V₂ - V₁) / t
Slope = Velocity / time ( which is the formula of acceleration)
Therefore, in figure a, the slope of the graph represent the acceleration of the object.
Therefore, take note that what ever the value that you obtained when calculating the slope is the acceleration of the object.
In figure b, the slope = 🔼V / 🔼t ➡ slope = change in velocity / change in time
Slope = ( Final velocity - initial velocity ) / time ➡ slope = (V₂ - V₁) / t
Slope = Velocity / time ( which is the formula of acceleration) . in figure b, V₂ & V₁ are the same because the velocity of the object is constant throughout the time frame. Therefore, V₂ = V₁. Slope = (V₂ - V₂) / t. ➡ slope = 0 / t. ➡ slope = 0.
Therefore, since slope = 0, acceleration of the object is equal to zero in velocity - time graph of uniform velocity. ( I.e acceleration = 0).
DECELERATION OR RETARDATION:
Retardation is the time rate of decrease in the velocity of an object. Deceleration or retardation is negative acceleration. The s.I.unit of retardation is m/s²
UNIFORM RETARDATION:
Uniform retardation is the constant time rate of decrease in the velocity of an object. The unit of retardation is m/s²
FORMULA FOR CALCULATING RETARDATION:
retardation = 🔼V / 🔼 t ➡ retardation = change in velocity / change in time.
➡ retardation = ( Final velocity - initial velocity ) / time ➡ retardation = (U - V) / t
You must take note that, retardation means reduction in velocity. In retardation, the initial velocity is always larger than the final velocity which in some cases are zero.
APPLICATION OF FORMULA :
EXAMPLE 1:
Attain decelerates from 25 m/s to 15 m/s in 2 minutes. Determine the retardation of the train.
SOLUTION:
Data given in the question:
Initial velocity U = 25 m/s , final velocity V = 15 m/s, time = 2 minutes = 2 * 60 seconds. Time = 120 seconds.
Forumla : retardation = ( V - U ) / time
Substitution: retardation = ( 15 - 25 ) / 120. ➡ retardation = - 10 / 120
➡ retardation = 0.0833 m/s² or acceleration = - 0.083 m/s²
Example 2:
A car which travels at the rate of 35 m/s slow down to a stop when the brakes where applied. If the time taken is 1½ minutes, calculate the deceleration seconds.
SOLUTION:
Data given in the question:
Initial velocity U = 35 m/s, final velocity V = 0 m/s ( because the car slow down and stop ), time = 1½ minutes, time = 1½ * 60 seconds = 3/2 * 60 seconds = 90 seconds
Formula: deceleration = ( V - U ) / time
Substitution: deceleration = ( 0 - 35 ) / 90 = ➡ deceleration = - 35 / 90
➡ deceleration = 0.388 m/s² or acceleration = - 0.388 m/s²
Example 3:
Given that the acceleration of a car is - 0.25 m/s² when the car slow down from its initial velocity Y m/s to 1.25 m/s in 35 seconds, calculate the initial velocity Y of the car.
SOLUTION:
Data given in the question:
Acceleration = - 0.25 m/s², initial velocity U = Y m/s, final velocity V = 1.25 m/s ,
time = 35 seconds.
Formula: deceleration = ( V - U ) / time
Substitution: - 0.25 = ( 1.25 - U ) / 35 ➡ - 0.25 * 35 = 1.25 - U. ➡ - 0.00714 = 1.25 - U
Make U the subject : U = 1.25 + 0.00714. ➡. U = 1.25714 m/s
EXERCISES:
VELOCITY-TIME GRAPH :
The velocity - time graph is obtained by plotting different values of velocity on the vertical axis against different values of time on the horizontal axis.
Velocity - time graph of constant retardation or deceleration
b. |
| velocity time graph for 1. constant acceleration, 2. Uniform velocity and 3. Constant deceleration |
AREA OF VELOCITY- TIME GRAPH:
From the figure above, the shape of the graph is the shape of a trapezium. You can see that the shape of the trapezium is made up of the shape of triangle + shape of rectangle + shape of triangle .
Shape of trapezium = shape of triangle + shape of rectangle + shape of triangle
Then,
Area of trapezium = area of triangle + area of rectangle + area of triangle
Area of trapezium = ½BH of triangle + L *B of rectangle + ½BH of triangle
Or , area of trapezium = ½ ( A+B ) * H of the tratravelled.
therefore,
Area of the velocity- time graph represent the distance travelled by object.
Area of velocity - time graph = distance travelled.
Note:
Since the velocity - time graph is made of a triangle, a rectangle and another triangle, the distances that the object travelled in each case of the motion can be calculated separately by calculating the area of each shapes in each time of the motion, using their formulae. This will be illustrated in the worked example below.
WORKED EXAMPLE:
A car start from rest and accelerated to 1.0 m/s for 2.5 seconds. The car then moved with uniform velocity for for 5 seconds. When the brakes where applied, the car slow down to a stop for 2.5 seconds.
When the car accelerates from rest to 1.0 m/s, U = 0 m/s, V = 1.0 m/s, time = 2.5Sec.
(I) acceleration, a = Velocity / time ➡ a = 1.0 / 2.5 ➡ a = 0.4 m/²
(ii)
When the brakes where applied,the car decelerate ( i.e slow down ) from 1.0 m/s to 0 m/s for 2.5 seconds. Then, U = 1.0 m/s, V 0 m/s, t = 2.5 seconds.
deceleration, d = change in velocity / time ➡ d = ( V - U ) / time
From the figure above, the shape of the graph is the shape of a trapezium. You can see that the shape of the trapezium is made up of the shape of triangle + shape of rectangle + shape of triangle .
Shape of trapezium = shape of triangle + shape of rectangle + shape of triangle
Then,
Area of trapezium = area of triangle + area of rectangle + area of triangle
Area of trapezium = ½BH of triangle + L *B of rectangle + ½BH of triangle
Or , area of trapezium = ½ ( A+B ) * H of the tratravelled.
therefore,
Area of the velocity- time graph represent the distance travelled by object.
Area of velocity - time graph = distance travelled.
Note:
Since the velocity - time graph is made of a triangle, a rectangle and another triangle, the distances that the object travelled in each case of the motion can be calculated separately by calculating the area of each shapes in each time of the motion, using their formulae. This will be illustrated in the worked example below.
WORKED EXAMPLE:
A car start from rest and accelerated to 1.0 m/s for 2.5 seconds. The car then moved with uniform velocity for for 5 seconds. When the brakes where applied, the car slow down to a stop for 2.5 seconds.
Calculate:
(i) the acceleration of the car. (ii) the deceleration of the car. (iii) the distance travelled by the car in each phase of the motion. (iv) the total distance travelled.
SOLUTION:
When the car accelerates from rest to 1.0 m/s, U = 0 m/s, V = 1.0 m/s, time = 2.5Sec.
(I) acceleration, a = Velocity / time ➡ a = 1.0 / 2.5 ➡ a = 0.4 m/²
(ii)
When the brakes where applied,the car decelerate ( i.e slow down ) from 1.0 m/s to 0 m/s for 2.5 seconds. Then, U = 1.0 m/s, V 0 m/s, t = 2.5 seconds.
deceleration, d = change in velocity / time ➡ d = ( V - U ) / time
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