## What is Acceleration?

In this article, we will discuss how to find acceleration. However, **we will start by defining** **Acceleration as the rate of change of velocity with respect to time**. Therefore, when an object undergoes acceleration, it can either speed up, slow down or change direction. Acceleration is a vector quantity, meaning it has both magnitude and direction. Additionally, It is denoted by the symbol “a” and is typically measured in meters per second squared (m/s²).

The magnitude of the acceleration is the change in velocity divided by the time it takes for the change to occur. The direction of acceleration is the same as the direction of the change in velocity.

### Causes of Acceleration

Acceleration can be caused by a number of factors, including gravity, friction, and a force applied to the object. For example, when an object is dropped, it accelerates downwards due to the force of gravity. Additionally, when an object is moving on a surface, it experiences friction, which opposes its motion. Thus, this frictional force causes the object to slow down. If a force is applied to an object in a certain direction, the object will accelerate in that direction.

The acceleration of an object can be calculated using the following equation:

a = Δv / Δt

- a is the acceleration
- Δv is the change in velocity
- Δt is the time it takes for the change in velocity to occur

where:

For example, if an object’s velocity changes from 5 m/s to 10 m/s in 2 seconds, its acceleration is 2.5 m/s².

Acceleration can be classified into two types: constant acceleration and variable acceleration.

**Constant acceleration**occurs when the acceleration of an object does not change over time. An example of constant acceleration is the free fall of an object.**Variable acceleration**occurs when the acceleration of an object changes over time. An example of variable acceleration is the motion of a car that is accelerating or decelerating.

## How to Find Acceleration: Formulas and Methods

For us to be able to find acceleration, we need to know the following acceleration formulas below:

- Acceleration (a) = Change in velocity (Δv) / time (Δt) or
**a = v / t** - a = (v – u) / t
- a = (v
^{2}– u^{2}) / 2s - a = F / m

Where: a = acceleration, v = final velocity, u = initial velocity, t = time, F = force, m = mass, and s = distance or displacement.

There are several methods to calculate acceleration, depending on the available information. We will now look into these methods and see how we can find acceleration:

### Method 1: Average Acceleration

To find the average acceleration of an object, you need to know the change in velocity and the time it takes. The formula for average acceleration is:

a = Δv / Δt

where:

- a represents acceleration
- Δv denotes the change in velocity
- Δt signifies the change in time

Therefore, by dividing the change in velocity by the change in time. You can easily determine the average acceleration of an object over a given time interval.

### Method 2: Instantaneous Acceleration

Instantaneous acceleration refers to the acceleration of an object at a specific moment in time. Thus, it involves calculating the derivative of velocity with respect to time

To find the instantaneous acceleration, you need to determine the derivative of velocity with respect to time. In other words, you calculate the rate at which velocity is changing at a specific instant. Additionally, the formula for instantaneous acceleration can be expressed as:

**The formula for instantaneous acceleration:**

a = dv/dt

From the formula above, “a” represents the instantaneous acceleration, “dv” represents the infinitesimal change in velocity, and “dt” represents the infinitesimal change in time. This method allows you to find the acceleration of an object at any given point in time.

### Method 3: Acceleration from Velocity-Time Graphs

Another way to find acceleration is by utilizing velocity-time graphs. If you have a graph that represents an object’s velocity as a function of time, you can determine the acceleration by examining the slope of the graph. The slope of a velocity-time graph corresponds to the object’s acceleration at a particular moment. A steeper slope indicates a higher acceleration, while a flatter slope represents a lower acceleration.

Remember that slope, s = y / x

and in the case of a velocity time graph, we have v representing y, while t is standing for x. We need to also remember that acceleration (a) is velocity (v) divided by time (t). In this case, s represents acceleration. Therefore, we can finally conclude that the velocity time graph can be written as: **a = v / t**.

## Solved Problems: How to Find Acceleration

To solidify your understanding of how to find acceleration, we are now going to solve few problems.

### Problem 1: Car Acceleration

Suppose a car initially at rest accelerates uniformly at a rate of 3 m/s² for 10 seconds. To find the final velocity, you can use the formula:

v = u + at

Where “v” represents the final velocity, “u” represents the initial velocity (which is 0 in this case), “a” represents the acceleration, and “t” represents the time.

And v = ?, u = 0, a = 3 m/s², and t = 10 s.

Substituting the above values into the equation (v = u + at), you will end up with:

v = 0 + (3 m/s²) x (10 s) = 30 m/s

**Therefore, the final velocity of the car after 10 seconds of acceleration is 30 m/s.**

### Problem 2: Free Fall

Consider an object in free fall near the Earth’s surface. The acceleration due to gravity, denoted by “g,” is approximately 9.8 m/s². If an object is dropped from a height, you can calculate the time it takes to reach the ground using the following equation:

h = (1/2) x g x t²

Where “h” represents the height, “g” represents the acceleration due to gravity, and “t” represents the time.

If the object is dropped from a height of 100 meters. By rearranging the equation and solving for “t,” you find:

t = √(2h / g)

Therefore

t = √(2h / g) = √(2 x 100 m / 9.8 m/s²) ≈ 4.52 s

**Hence, the object takes approximately 4.52 seconds to reach the ground.**

### Problem 3: Deceleration

A particle decelarates uniformly from rest at 6.0 m/s² for 8 seconds and then decelarates uniformly to rest in the next 5 seconds. Determine the magnitude of the deceleration.

**Data:**

Initial velocity, u = 0,

acceleration, a = 6.0 m/s²,

Time, t = 8 s,

final velocity, v = ?

We will now use the equation, v = u + at = 0 + 6 x 8 = 48 m/s

Therefore, the particle decelerated to rest from a velocity of 48 m/s. Now we can calculate the deceleration in this way:

u = 48 m/s, v = 0, t = 5 s, deceleration (a) = ?

By applying the same equation, which is v = u + at. We will now have

0 = 48 + a x 5

After making a subject of the formula, we will have

a = – 48 / 5 = – 9.6 m/s² [Note: the negative sign indicates deceleration]

We can now remove the negative sign and show that deceleration (a) is 9.6 m/s²

a = 9.6 m/s²

**Therefore, the particle decelerates at 9.6 meters per second square.**

### Problem 4

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

**Data**: Information from the question

Initial velocity, u = 0

Final veloicty of the motorcycle, v = 108 km/h = (108 x 1000) / (60 x 60) = 30 m/s

Time, t = 15 s

We will now use the formula for acceleration which **a = (v – u) / t**

Thus, a = (v – u) / t = (30 – 0) / 15 = 30 / 15 = 2 m/s²

**Therefore, the acceleration of the motorcycle is 2 meters per second square.**

### Problem 5

A body accelerates uniformly from rest at 6ms^{−2} for 8 seconds and then decelerates uniformly to rest in the next 5 seconds. Determine the magnitude of the deceleration.

**Solution**

**Data:** The information from the question

Initial velocity (u) = 0, acceleration (a) = 6ms^{−2}, time (t) = 8s, velocity attained (v) = ?

**Formula: ** The equation that will help us solve the problem

v=u+at

Now, substituting the data into the above formula, we will have

v = 0+6×8 = 48 m/s

We will now find the deceleration as the particle decelerated from 48m/s to rest. Thus;

u = 48m/s, v = 0, t = 5s, decelaration (a) = ?

Now, v=u+at

Hence, 0 = 48 + a x 5

By making a subject of the formula, we will now obtain

a = – 48/5 = – 9.6 m/s^{2}

**The negative sign in the above answer is the sign of deceleration. Thus, the deceleration (a) is 9.6 m/s ^{2} **

### Problem 6

The velocity v of a space shuttle in a time t is given by v=25+3t^{2}. Find the instantaneous acceleration at t = 9s.

#### Solution

a = dv/dt

Since, v=25+3t^{2}

differentiating the above equation with respect to t, will give us:

dv/dt = 0 + 2(3t^{2-1}) = 6 x t = 6t

and t=9s

**Therefore, dv/dt = 6 x 9 = 54 m/s**.

## Frequently Asked Questions

### 1. What is the unit of acceleration?

The unit of acceleration is meters per second squared (m/s²). This means that acceleration is the change in velocity per unit time. For example, if an object’s velocity changes from 5 m/s to 10 m/s in 2 seconds, its acceleration is 2.5 m/s².

### 2. Can acceleration be negative?

Yes, acceleration can be negative. A negative acceleration indicates that an object is slowing down or changing direction in the opposite sense of its initial motion.

### 3. How is acceleration different from velocity?

Velocity represents the rate at which an object’s position changes, while acceleration measures how quickly an object’s velocity is changing.

### 4. What is the relationship between acceleration and force?

According to Newton’s second law of motion, the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. The formula is given as:

### 5. What is the difference between acceleration and velocity?

Velocity is the rate at which an object changes its position, while acceleration is the rate at which an object changes its velocity. Velocity has both magnitude and direction, while acceleration only has magnitude.

### 6. What are some examples of acceleration?

Some examples of acceleration include:

- The free fall of an object
- The motion of a car that is accelerating or decelerating
- The motion of a ball that is thrown into the air
- The motion of a planet orbiting a star

### 7. What are the effects of acceleration on an object?

The effects of acceleration on an object include:

- A change in the object’s velocity
- A change in the object’s momentum
- A change in the object’s kinetic energy
- A force being exerted on the object

### 8. How to find acceleration with mass and force?

Since Force(f) = Mass(m) x Acceleration(a). We will use the formula a = f / m. Where: a = acceleration, f = force, and m = mass.

### 9. How to find acceleration given time and distance?

To find acceleration given time and distance, we use the formula for acceleration: **a = 2s / t ^{2}, **where the initial velocity is zero (starting from rest). Thus, u = 0, s = distance covered, and t is the time taken.

### 10. How to find acceleration with velocity and time?

To find acceleration with velocity and time, we will use the formula **a = v / t**. Where a = acceleration, v = velocity, t = time taken.

### 11. How to find acceleration without time?

To find acceleration without time, we will have to use the formula a = f / m. Where a = acceleration, f = force, and m = mass.

### 12. How to find acceleration with velocity and distance?

To find acceleration with velocity and distance, we will apply one of the formulae for kinematic equations that says v^{2} = u^{2} + 2as. Therefore, by making a subject of the formula, we will have: a = (v^{2} – u^{2}) / 2s which is the formula for finding acceleration with velocity and distance.

### 13. How to find acceleration due to gravity?

To find acceleration due to gravity, we will apply the formula that says g = GM / r^{2}. Where g = acceleration due to gravity = 9.8 m/s^{2}, G = gravitational constant = 6.68 x 10^{-11} m^{3}kg^{-1}s^{-2}, r = radius of the earth = 6,371 km, and M = mass of the earth = 5.972 × 10^{24} kg

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