Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing.
Acceleration can also be termed as the rate of change in velocity and the change over time. An acceleration vector’s magnitude tells us how much the velocity will change, while its direction tells us how the velocity will change i.e. whether the velocity is increasing or decreasing, the velocity vector is changing direction, or some combination of the three.
Acceleration can be positive, zero, or negative. In case the object’s velocity increases with time, it can be termed Positive Acceleration. In case the velocity is zero, it is termed Zero Acceleration, while the negative acceleration also known as retardation indicates a decrease in velocity with time.
Acceleration Formula
Mathematically, the change in the velocity of an object in motion is defined as, (v – u) where v and u are the final and the initial velocities.
Therefore, the acceleration of the object is given by,
Acceleration = Change in Velocity / Time Taken
a = (v – u) / t
where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time taken by the object
Unit of Acceleration
It is a vector quantity, which is associated with both magnitude and direction. It is denoted by ‘a’.
The unit of acceleration is meters per second squared or meters per second (the object’s speed or velocity) per second or m/s2.
Note: Dimensional Formula of Acceleration is [M0 L1 T-2]
Types of Acceleration
Following are the different types of acceleration associated with an object,
Uniform Acceleration
Non-Uniform Acceleration
Average Acceleration
Instantaneous Acceleration
Now let’s learn about each type of acceleration in detail.
Uniform Acceleration
In case the velocity of an object changes in equal amounts during the same time interval, then the body is said to be in uniform acceleration. In this case, neither the direction nor magnitude changes with respect to time.
For Examples:
A ball rolling down the slope.
When a bicycle rider is riding the bicycle on a slope where both pedals are engaged.
A kid sliding down from the slider.
Motion of car with constant velocity, etc
Non-Uniform Acceleration
When the velocity of the particle is constant or acceleration is zero.
When the particle is moving with a constant acceleration and its initial velocity is zero.
When the particle is moving with constant retardation.
When the particle moves with non-uniform acceleration and its initial velocity is zero.
When the acceleration decreases and increases.
The total area enclosed by the time-velocity curve represents the distance travelled by a body.
Difference Between Acceleration and Velocity
Solved Examples on Acceleration
Example 1: If a truck accelerates from 6 m/s to 10 m/s in 10 s. Calculate its acceleration.
Solution:
Given that,
Initial Velocity, u = 6 m/s,
Final Velocity, v = 10 m/s,
Time taken, t = 10 s.
We have to find Acceleration ‘a’
Acceleration, a = (v – u) / t
= (10 m/s – 6 m/s) / 10 s
= 0.4 m/s2
Thus, the acceleration of the truck is 0.4 m/s2.
Example 2: If a ball is released from the terrace of a building to the ground. If the ball took 6 s to touch the ground. Find the height of the terrace from the ground.
Solution:
Given that,
Initial Velocity u = 0 {as the ball was at rest},
Time taken by the ball to touch the ground t = 6 seconds
Acceleration due to gravity a = g = 9.8 m/s2,
Distance traveled by stone = Height of bridge = s
The distance covered by the ball from the terrace to the ground
Therefore,
Distance of the terrace from the ground is 29.4 m.
Example 3: If a man is driving the car at 108 km/h slow down and bring it to 72 km/h in 5 s. Calculate the retardation of the car?
Solution:
Given that,
Initial velocity, u = 108 km/h or
Final velocity, v = 72 km/h or
Time taken, t = 5 seconds
Therefore, acceleration is,
The negative sign shows retardation.
Example 4: If a car moves from rest and then accelerates uniformly at the rate of 7.5 m/s2 for 10 s. Find the velocity of the train in 10 s.
Solution:
Given that,
Initial velocity u = 0 {as the car was at rest}
Acceleration a = 7.5 m/s2
Time t = 10 s
v = u + at
= 0 + 7.5 × 10
= 75 m/s
Example 5: If an object moves along the x-axis according to the relation x = 1 – 2t + 3t2, where x is in meters and t is in seconds. Calculate the acceleration of the body when t = 3s.
Solution:
Given that,
x = 1 – 2t + 3t2
Velocity, v = dx/dt
= d/dt {1 – 2t + 3t2}
= -2 + 6t
Therefore, Acceleration a = dv/dt
= d/dt {-2 + 6t}
= 6 m/s2
FAQs on Acceleration
Question 1: What is Radial Acceleration?
Answer:
The acceleration of an object along the radius towards the center of the radial path is called the radial acceleration.
Question 2: What is Acceleration due to Gravity?
Answer:
The acceleration due to gravity is defined as the acceleration experienced by the earth’s gravitational pull. It value is 9.8 m/s2
Question 3: What is Tangential Acceleration?
Answer:
Tangential acceleration is the rate at which a tangential velocity varies in the rotational motion of any object. It acts in the direction of a tangent at the point of motion for an object.
Question 4: What do you understand by Centripetal Acceleration?
Answer:
Centripetal Acceleration is defined as the acceleration points towards the center of the curvature, now since the velocity is continuously changing and acceleration is present.
Question 5: How to find the Acceleration?
Answer:
The formula to find the acceleration of any object is,
a = (v – u) / t
where, v is final velocity, u is initial velocity, t is the time taken
Question 6: What is the SI unit of Acceleration?
Answer:
The SI unit of the accerelation is m/s2
Question 8: What is Uniform Acceleration?
Answer:
When an object accelerates uniformly i.e. its velocity increases linearly with time then it is called to be in uniform acceleration
Question 9: What is the Acceleration of any object in Free Fall?
Answer:
In free fall the acceleration of an object is also called the acceleration due to gravity and is equal to 9.8m/s2
A+0+70
ACCELERATION
Film Loop: Inertial Forces – Centripetal Acceleration
Length: 3:15 Min., Black and White, No Sound
This film loop was made at the Rotor Ride at Cedar Point, Sandusky, Ohio. The cylindrical rotating device has inside diameter 14 ft. and attains a maximum angular speed of 27 rev/min. From these data, the centripetal acceleration is 56 ft/sec squared, about 1.8 g. After full speed is reached, the floor drops down and the passengers remain affixed to the wall.
There are two equivalent ways of analyzing the situation. To an outside observer, the rotation is known to exist (relative to a inertial frame). The wall supplies an inward and upward force P; the resultant of this force and the weight mgis horizontal and is the centripetal force which causes the centripetal acceleration. The rider’s outward force on the wall is the reaction to the force of the wall on the rider, and this force is not shown in the diagram because it acts on the wall, not the rider. An upward component of the force of the wall on the rider (friction) arises because of the normal component of the force between the rider and the wall.
From an insider’s point of view, an outward inertial force has come into existence because of the acceleration of his frame of reference. This outward force can properly be called “centrifugal force” by an observer who is in the accelerated frame of reference. This force is identical in its effect to an outward gravitational force; it is “artificial gravity ” of magnitude 1.8g. The rider considers himself to be in equilibrium under the action of three forces; P, mg, and the inertial force- ma. .
In the film, the camera views the action from both frames of reference. (The cameraman hand-holds the camera while enjoying the ride. No special support is used.) Viewed from inside the Rotor the resultant gravity is downward and outward, as shown by the beach ball which no longer hangs vertically.
The interior wall of the Rotor is of heavy padded fabric rough enough to supply the necessary friction. The coefficient of friction between this fabric and ordinary clothing is evidently somewhat greater than 0.55.
A+0+75
ACCELERATION
Film Loop: Inertial Forces – Translational Acceleration
Length: 2:05 min., Black and White, No sound
A 156-lb student riding in an elevator experiences an increase in weight when the elevator starts up and a decrease when it accelerates downward. When moving at constant speed (between floors) his weight is normal.
The elevator in the Buckeye Federal Savings and Loan building in Columbus, Ohio, was selected for its large acceleration and relatively smooth stop. The safety interlock is disabled so that the elevator can be operated with its doors open. In this way the direction of motion can be seen as the floors go by. The indicator of the spring scale overshoots the mark; the actual increase and decrease of weight is somewhat less than the maximum readings of the indicator.
There are two equivalent ways of analyzing the forces. To an outside observer the acceleration is known to exist (relative to an inertial frame). The push of the floor on the student is P. The resultant force is P-mg, and Newton’s 2nd law says P-mg=ma, whence P = m (g+a). For an upward acceleration a>0, P>mg, and the floor pushes upward with a force greater than the student’s normal weight. According to Newton’s 3rd law, the student pushes downward on the floor with a force of magnitude P, and therefore the scale registers the force Pwhich is greater than mg. Similarly when a is negative, the apparent weight is less than mg.
If the student does not know the elevator is accelerating, he considers himself to be in equilibrium under the action of two forces: the push of the scale platform P, and a “gravitational” force -m (g+a). The inertial force -ma which has arisen because of the (unknown to him) acceleration of his frame of reference is in every respect equivalent to a gravitational force. He is at liberty to say either “someone accelerated the elevator upward” or “someone turned on an extra downward gravitational force.”
Let’s now discuss some important terms related to a Simple Harmonic Motion of a particle as
Displacement (x): Displacement at any instant of time is defined as the net distance travelled by the body executing SHM from its mean or equilibrium position.
Amplitude (A): The amplitude of oscillation is defined as the maximum displacement of the body executing SHM on either side of the mean position.
Velocity (v): Velocity at any instant is defined as the rate of change of displacement with time. For a body executing SHM, its velocity is maximum at the mean position and minimum (zero) at the extremes. The Velocity of the body is inversely proportional to the displacement from the mean position.
Acceleration (a): Acceleration is defined as the rate of change of velocity with time. Unlike velocity, acceleration is directly proportional to displacement. It is maximum at the extreme positions where the displacement is maximum and minimum at the mean position (displacement = 0).
Restoring Force (FR): Restoring Force is the force that always acts in a direction opposite to that of displacement but is directly proportional to it. Restoring Force is maximum at the extreme positions and minimum at the mean position.
Spring Constant (k): Spring Constant is a constant value for a particular spring that determines the amount of force required to compress or stretch the spring by 1 unit.
Energy (E): The total energy of the body under SHM is called mechanical energy, mechanical energy of the body remains constant throughout the motion if the medium is frictionless. The Mechanical Energy of a body at any instant is the sum total of its kinetic and potential energy.
Time Period (T): The Time Period of oscillation is defined as the time taken by the body to complete one oscillation. In other words, it is the time taken to cover 4 times the amplitude.
Frequency (f): Frequency is defined as the number of oscillations made by the body in one second. It is reciprocal of the time period. f = (1/T)
