Acceleration - Examples, Meaning, Types, Formula, Faqs

As we shall study, the acceleration of an object is the change in its velocity in each unit of time. In case the change in velocity in each unit of time is constant, the object is said to be moving with constant acceleration and such a motion is called uniformly accelerated motion. On the other hand, if the change in velocity in each unit of time is not constant, the object is said to be moving with variable acceleration and such a motion is called non-uniformly accelerated motion.

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This Story also Contains

1. What is Acceleration?
2. Explain Acceleration with An Example.
3. What is the Meaning of Acceleration?
4. Positive, Negative and Zero Acceleration
5. Solved Examples Based on Acceleration

This concept belongs to the chapter Kinematics, which is an important chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), almost seven questions have been asked on this concept. And for NEET three questions were asked from this concept.

What is Acceleration?

The rate at which a person's velocity changes in both speed and direction over time is called acceleration. A point or object is accelerated if it moves faster or slower in a straight line. On a circle, the direction of motion is constantly changing. Even if the speed is constant, the motion is accelerated. Both effects contribute to the acceleration of all other types of motion.

$ \vec{a}=\frac{\text { change in velocity }}{\text { time taken }}=\frac{\vec{v}_f-\vec{v}_i}{t}$

Where: $\overrightarrow{v_f}=\text { Final velocity, } \vec{v}_i=\text { initial velocity and } \mathrm{t}=\text { time }$

Tips for Acceleration

1. The body is said to have undergone acceleration if there is a change in velocity i.e.,

   • Change in speed

   • Change in direction

   • Change in both

2. It is a vector quantity

3. $\text { Dimension }=L T^{-2}$

4. $\text { S.I unit }=m s^{-2}$

Explain Acceleration with An Example.

Let us understand what is the meaning of acceleration with the help of some examples.

When we are on a roller coaster ride, as it starts, we experience a push backwards against our seats, and when it stops, we experience a push forward, and whenever it makes a sharp turn, we experience a push sidewards. This push experienced is all because of acceleration.

When the ride starts or stops, there is a change in its speed. And when it makes a turn, there is a change in its direction.

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What is the Meaning of Acceleration?

Acceleration meaning is whenever a body changes its speed or direction in a motion.

Or, there will be acceleration whenever there is a change in velocity.

Acceleration Formula in General

We are familiar that velocity is a vector quantity because it is a speed with direction. The acceleration 'a' is calculated as follows:

a = Velocity Change/Time Taken

This formula indicates that the acceleration is the rate of change in velocity, or if an object's velocity changes from its initial value "u" to its final value "v," the statement can be expressed as:

(v - u)/t = a

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Physics' Acceleration Formula

Define acceleration in physics.

Acceleration in physics is well-defined as the rate at which an object's velocity changes, regardless of whether it speeds up or slows down. If it accelerates, the acceleration is positive; if it slows, the acceleration is negative. According to Newton's Second Law, it is caused by the object's net imbalanced force. Because it describes the rate of velocity change with respect to time, which is a vector quantity, acceleration is a vector quantity. The letter a stands for acceleration. It has the SI unit of m/s² and the dimensions M⁰L¹T^-2.

If v_o denotes initial velocity, v_t denotes final velocity, and t denotes time taken.

Acceleration is equal to:

a=(v_t-v₀)/t

Instantaneous acceleration formula:

If r denotes displacement vector,
v=ⅆr/ⅆt denotes velocity

Acceleration is equal to,
a=ⅆv/ⅆt=ⅆ²r/ⅆt²

Types of Acceleration

1. Uniform acceleration

2. Non-uniform acceleration

Define Uniform acceleration

We say an object has uniform acceleration if its speed (velocity) increases at a constant pace. That is, there is no change in the rate of acceleration.

Let us understand uniform acceleration thoroughly.

As the name implies, uniformly accelerated motion refers to an object or a body that accelerates at a consistent rate. Constant velocity does not imply constant acceleration. The definition of uniform acceleration is when a body is in motion, and the amount of variation in velocity in equal intervals of time is constant.

Example of Uniform acceleration

For an example of uniform acceleration, consider the motion of a freely falling body, where the body's acceleration is the sole acceleration attributable to gravity. When we plot velocity vs time on a graph, we get a straight line, and the entire slope equals the required acceleration. Real-life example would be when someone is parachute diving.

Define non-uniform acceleration

Non-uniform acceleration occurs when an object's velocity varies in variable amounts during equal time intervals.

The opposite of uniform acceleration is non-uniform acceleration. We know that uniform acceleration indicates that the rate of change in velocity remains constant across time. In a non-uniform acceleration, the change in velocity is not the same. The magnitude of acceleration and the direction of velocity will change over time.

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Example of non-uniform acceleration

Imagine driving a car on the road; there is a recurrent increase or decrease in the vehicle's velocity at unequal intervals of time, causing the vehicle to experience non-uniform acceleration.

Positive, Negative and Zero Acceleration

• When the velocity of the body increases with time, it experiences positive acceleration. It means the velocity-time graph has a positive slope.

• When the velocity of a body decreases with time, it experiences negative acceleration. It means the velocity-time graph has a negative slope.

• When the velocity of a body is constant, or the body is at rest, it experiences zero acceleration. It means the velocity-time graph slope is zero.

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Solved Examples Based on Acceleration

Example 1: The displacement Vs time graph is given below which of the following conclusions is correct:

1) Velocity is constant
2) Velocity is continuously increasing

3) Acceleration is positive

4) Acceleration is negative

Solution :

As we know the slope of the displacement-time graph gives us velocity. Hence, a change in slope represents a change in velocity.

In the above figure, the velocity of the particle i.e. slope in the displacement vs time graph is decreasing.

Therefore, acceleration is negative.

Hence, the answer is option (4).

Example 2: Three position Vs time graph is shown in the figure. In which case acceleration is zero

1) I

2) II

3) III

4) None of these

Solution:

Zero Acceleration

When the final velocity is equal to the initial velocity.

For: V₁ = V₂

wherein

When a bus is moving with uniform velocity.

Since velocity is constant in case II hence acceleration = 0

Hence, the answer is option (2).

Example 3: Four velocity time graphs (Namely I, II, III, IV ) are shown in the figure. In which case is the acceleration uniform and positive?

Solution:

$ a=\frac{d v}{d t}$= slope in the V - t graph.

As it is clear II has a uniform and positive slope so it indicates uniform and positive acceleration.

Hence, the answer is option (2).

Example 4: The position of particles as a function of time $t$ is given by $x(t)=a t+b t^2-c t^3$ where $\mathrm{a}, \mathrm{b}$, and $\mathrm{c}$ are constants. When the particle attains zero acceleration, then its velocity will be :

1) $a+\frac{b^2}{4 c}$
2) $a+\frac{b^2}{3 c}$
3) $a+\frac{b^2}{c}$
4) $a+\frac{b^2}{2 c}$

Solution :

Given:

$\begin{aligned}
& x(t)=a t+b t^2-c t^3 \\
& v=\frac{d x}{d t}=a+2 b t-3 c t^2 \\
& a=\frac{d v}{d t}=2 b-6 c t \\
& a=0 \Rightarrow t=\frac{2 b}{6 c}=\frac{b}{3 c}
\end{aligned}$

$\begin{aligned}
& \text { So, Velocity at time } t=\frac{b}{3 c} \\
& V=a+2 b \times \frac{b}{3 c}-3 c \times \frac{b^2}{9 c^2} \\
& V=a+\frac{b^2}{3 c}
\end{aligned}$

Hence, the answer is the option (2).

NCERT Physics Notes :

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