How to Use the Calculator?
To use this calculator, first choose whether you want to calculate acceleration from speed difference or distance traveled. Then enter any three values you know, such as initial speed, final speed, distance, time, or acceleration. The calculator will automatically fill in the missing value as soon as enough information is entered.
You can also change the units next to each field, and the calculator will convert the value for you.
What Is Acceleration?
Acceleration tells us how quickly an object’s velocity changes. In everyday language, we often say an object is accelerating when it is “speeding up,” but in physics, acceleration has a wider meaning.
An object accelerates whenever its velocity changes in any way. This means it can speed up, slow down, or change direction.
Velocity means speed in a particular direction. If a car starts from rest and reaches a speed of $20 \text{ m/s}$ in $5 \text{ s}$, its velocity has changed. That change did not happen instantly; it happened over time. Acceleration describes the rate of that change.
The most common formula for acceleration is:
$$a=\frac{v_f-v_i}{t}$$
In this formula, $a$ means acceleration, $v_f$ means final velocity, $v_i$ means initial velocity, and $t$ means time. The numerator $v_f-v_i$ is called the change in velocity.
Let’s take an example.
If a cyclist increases speed from $4 \text{ m/s}$ to $10 \text{ m/s}$ in $3 \text{ s}$, the velocity has increased by $6 \text{ m/s}$. Since this increase happens over $3 \text{ s}$, the acceleration is $2 \text{ m/s}^2$.
$$ \begin{aligned} a &= \frac{10 – 4}{3} \\[15pt] a &= \frac{6}{3} \\[15pt] a &= 2 \text{ m/s}^2 \end{aligned} $$
This answer means the cyclist’s velocity increases by $2 \text{ m/s}$ every second. It does not mean the cyclist travels $2 \text{ m}$ every second. That would be speed. Acceleration is about how the speed or velocity changes.
Notice that the standard unit of acceleration is metres per second squared, written as $\text{m/s}^2$. This unit may look strange at first, but it has a clear meaning. If an object has an acceleration of $2 \text{ m/s}^2$, it means its velocity changes by $2 \text{ m/s}$ every second. So after one second, it is $2 \text{ m/s}$ faster; after two seconds, it is $4 \text{ m/s}$ faster; after three seconds, it is $6 \text{ m/s}$ faster, assuming the acceleration stays constant.
Positive, Negative, and Zero Acceleration
Acceleration can be positive, negative, or zero.
Positive Acceleration
Positive acceleration happens when the final velocity is greater than the initial velocity. For example, a running man may increase his speed from $3 \text{ m/s}$ to $7 \text{ m/s}$. Since his final velocity is greater than his initial velocity, he is speeding up, so his acceleration is positive.
Negative Acceleration
Negative acceleration happens when the final velocity is smaller than the initial velocity. For example, a boy on skates may slow down from $8 \text{ m/s}$ to $2 \text{ m/s}$. Since his velocity is decreasing, the value of $v_f-v_i$ becomes negative, so the acceleration is negative. This means he is slowing down.
Zero Acceleration
Zero acceleration happens when the velocity does not change. For example, if a car moves at a constant velocity of $15 \text{ m/s}$ for $10 \text{ s}$, then $v_f$ and $v_i$ are both $15 \text{ m/s}$. The difference is $0$, so the acceleration is $0 \text{ m/s}^2$. The car is still moving, but it is not accelerating because its velocity is not changing.
Understanding Distance Travelled with Acceleration
Sometimes we do not know the final velocity, but we do know the distance travelled, the initial velocity, and the time. In that case, we can find acceleration using a formula based on distance. This formula comes from the motion equation for constant acceleration.
$$d=v_i t+\frac{1}{2}at^2$$
This equation says that the total distance travelled comes from two parts. The first part, $v_i t$, is the distance the object would travel if it kept moving at its initial velocity without speeding up or slowing down. The second part, $\frac{1}{2}at^2$, is the extra distance caused by acceleration.
To solve this equation for acceleration, we rearrange it. First subtract $v_i t$ from both sides, and then multiply by $2$ and divide by $t^2$. The result is:
$$a=\frac{2(d-v_i t)}{t^2}$$
In this formula, $d$ means distance travelled, $v_i$ means initial velocity, $t$ means time, and $a$ means acceleration.
Imagine a skateboarder starts with an initial velocity of $2 \text{ m/s}$ and travels $30 \text{ m}$ in $5 \text{ s}$. We want to find the acceleration. Here, $d=30 \text{ m}$, $v_i=2 \text{ m/s}$, and $t=5 \text{ s}$.
$$ \begin{aligned} a &= \frac{2(d – v_i t)}{t^2} \\[15pt] a &= \frac{2(30 – 2 \cdot 5)}{5^2} \\[15pt] a &= \frac{2(30 – 10)}{25} \\[15pt] a &= \frac{40}{25} \\[15pt] a &= 1.6 \text{ m/s}^2 \end{aligned} $$
The skateboarder’s acceleration is $1.6 \text{ m/s}^2$. This means the skateboarder’s velocity increases by $1.6 \text{ m/s}$ every second.
Solved Examples
Let us try several examples.
Example 1: Finding Acceleration from Initial and Final Velocity
Problem: A car starts from rest and reaches a velocity of 20 m/s in 10 s. Find its acceleration.
Solution: The phrase “starts from rest” means the initial velocity ($v_i$) is 0 m/s. The final velocity ($v_f$) is 20 m/s, and the time ($t$) is 10 s. Since we are given initial velocity, final velocity, and time, we should use the formula involving velocity difference:
$$ \begin{aligned} a &= \frac{v_f – v_i}{t} \\[15pt] a &= \frac{20 – 0}{10} \\[15pt] a &= 2 \text{ m/s}^2 \end{aligned} $$
The car’s acceleration is 2 m/s². This means that every second, the car’s velocity increases by 2 m/s. Since the answer is positive, the car is speeding up.
Example 2: Finding Acceleration When an Object Slows Down
Problem: A bicycle is moving at 18 m/s. The rider applies the brakes, and the bicycle slows down to 6 m/s in 4 s. Find the acceleration.
Solution: Here, the initial velocity is 18 m/s, the final velocity is 6 m/s, and the time is 4 s. Since the final velocity is smaller than the initial velocity, we should expect a negative answer.
$$ \begin{aligned} a &= \frac{v_f – v_i}{t} \\[15pt] a &= \frac{6 – 18}{4} \\[15pt] a &= \frac{-12}{4} \\[15pt] a &= -3 \text{ m/s}^2 \end{aligned} $$
The acceleration is -3 m/s². The negative sign tells us that the bicycle is slowing down. In simple words, the bicycle loses 3 m/s of velocity every second.
Example 3: Finding Acceleration Using Distance Travelled
Problem: A motorbike has an initial velocity of 4 m/s. It travels 60 m in 5 s with constant acceleration. Find its acceleration.
Solution: In this problem, we are given distance, initial velocity, and time. We are not directly given final velocity, so we should use the distance formula for acceleration.
$$ \begin{aligned} a &= \frac{2(d – v_i t)}{t^2} \end{aligned} $$
Now substitute $d = 60 \text{ m}$, $v_i = 4 \text{ m/s}$, and $t = 5 \text{ s}$.
$$ \begin{aligned} a &= \frac{2(60 – 4 \times 5)}{5^2} \\[15pt] a &= \frac{2(60 – 20)}{25} \\[15pt] a &= \frac{2(40)}{25} \\[15pt] a &= \frac{80}{25} \\[15pt] a &= 3.2 \text{ m/s}^2 \end{aligned} $$
The motorbike’s acceleration is 3.2 m/s². This means the motorbike’s velocity increases by 3.2 m/s every second. The answer is positive because the motorbike travelled farther than it would have if it had continued at only 4 m/s.
Example 4: Checking Whether Acceleration Is Positive or Negative from Distance
Problem: A small cart is moving with an initial velocity of 10 m/s. It travels 35 m in 5 s. Find its acceleration.
Solution: First, think about what would happen if the cart kept moving at 10 m/s for 5 s without accelerating. It would travel:
$$ \begin{aligned} v_i t &= 10 \times 5 \\[15pt] v_i t &= 50 \text{ m} \end{aligned} $$
But the cart actually travels only 35 m. This is less than 50 m, so the cart must have slowed down. That means we should expect a negative acceleration.
$$ \begin{aligned} a &= \frac{2(d – v_i t)}{t^2} \\[15pt] a &= \frac{2(35 – 10 \times 5)}{5^2} \\[15pt] a &= \frac{2(35 – 50)}{25} \\[15pt] a &= \frac{2(-15)}{25} \\[15pt] a &= \frac{-30}{25} \\[15pt] a &= -1.2 \text{ m/s}^2 \end{aligned} $$
The cart’s acceleration is -1.2 m/s². The negative sign shows that the cart is slowing down. This makes sense because the cart travelled less distance than it would have travelled at its starting velocity.
