In physics, **acceleration** is the name given to a vector magnitude (it has direction) that indicates the variation in speed of a moving body as time passes. It is normally represented by the letter a and its unit of measurement, in the International System it is meters per second squared (m / s ^{2} ).

The origin of this concept comes from Newtonian mechanics, whose postulates assure that an object will always move in a rectilinear way unless forces that lead to acceleration impinge on it, resulting in a curved path.

This means that an object in rectilinear motion can only vary its speed if a force acting on it causes an acceleration: in the same direction of its trajectory (acceleration: it gains speed) or in the opposite direction of it (deceleration: it loses speed) .

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### Formula to calculate acceleration

Thus, classical mechanics defines acceleration as the variation of speed over time and proposes the following formula:

a = dV / dt

Where *a* will be acceleration, *dV* the difference of the speeds and *dt* the time in which the acceleration occurs. Both variables are defined as follows:

dV = V _{f} – V _{i}

Where *V _{f}* will be the final speed and

*V*the initial speed of the mobile. It is important to observe this order to reflect the direction of acceleration: there can be either a positive (gain speed) or a negative (lose speed) acceleration. Plus:

_{i}dt = tf – ti

Where *t _{f}* will be the final time and

*t*the initial time of the movement. Unless otherwise stated, the initial time will always be 0 seconds.

_{i}### Acceleration in relation to force

On the other hand, Newtonian mechanics establishes for a body of constant mass (m), contemplated by an inertial observer, a relationship of proportionality with respect to the force applied to the object (F) and the acceleration obtained (a), that is:

F = m. to

This relationship is valid in any inertial reference system and allows the acceleration to be calculated with the following formula:

a = F / m

This formulation obeys Newton’s Second Law (Fundamental Law of Dynamics).

### Acceleration calculation examples

- A racing car increases its speed at a constant rate of 18.5 m / s at 46.1 m / s in 2.47 seconds. What will its average acceleration be?

a = dv / dt = (v _{f} – v _{i} ) / (t _{f} – t _{i} ) where v _{f} = 46.1 m / s, v _{i} = 18.5 m / s, t _{f} = 2.47 s , t _{i} = 0 s.

So: a = (46.1 – 18.5) / 2.47 = **11.17 m / s ^{2}**

- A motorcyclist travels at 22.4 m / s and realizes that he has mistaken the route. Apply the brakes and the motorcycle stops in 2.55 s. What will be its slowdown?

a = dv / dt = (v _{f} – v _{i} ) / (t _{f} – t _{i} ), where v _{f} = 0 m / s, v _{i} = 22.4 m / s, t _{f} = 2.55 s, t _{i} = 0 s.

So: **a = (0 – 22.4) / 2.55 = -8.78 m / s ^{2
}**

- A force of 10 newtons of magnitude acts uniformly on a mass of 2 kilograms. What will be the acceleration of the pushed object?

a = F / m, where F = 10 N, m = 2Kg.

Thus:

**a = 10/2 = 5 m / s ^{2}**

- Someone is pulling a 400 kg piece of furniture to one side with a net force of 150 newtons. Another person pushes it in the same direction with a force of 200 newtons, but there is a wind blowing in the opposite direction with a force of 10 newtons. What will be the acceleration of the furniture?

We know that a = F / m, where the net force will be the sum of those in the same sense minus the one that opposes them: F = 150 N (person 1) + 200 N (person 2) – 10N (wind), which results in 340 N. We also know that m = 400 kg.

Then: **a = 340 N / 400 kg = 0.85 m / s ^{2}**

- A remote-controlled aircraft, with a mass of 10 kg, flies with an acceleration of 2 m / s
^{2}heading North. Just then a wind blows to the east, with a force of 100 N. What will be the new acceleration of the airplane that maintains its course?

Since the force of the wind is perpendicular to the direction of movement of the airplane, it will not have any effect on its movement. It will continue accelerating north at 2 m / s ^{2} .

- Two boys, one weak and one strong, play tug-of-war, each at one end of the rope. The first pulls to the left does so with a force of 5 N and the second pulls in the opposite direction does so with a force of 7 N. Taking into account that 1 newton (N) is equal to 1 kilogram-meter / second squared ( kg-m / s2), what will be the acceleration reached by the body of the weakest child, when being dragged in the opposite direction by the other?

From F = ma we know that a = F / m, but we must find the net force, which will be 2 N (7 N for the strong child – 5 N for the weak child).

Then, we must find the mass, which for the purposes of the calculation must be derived from the force that the weak child opposes, namely: 1 N = 1kg.m / s ^{2} , that is, it is the amount of force to mobilize a kilogram of mass at one meter per second squared.

Therefore, 5N = 5kg.m / s ^{2} . Hence m = 5Kg.

**And finally, we know that a = 2N (F) / 5kg (m) = 0.4 m / s ^{2}**

- A fire truck increases its speed from 0 to 21m / s to the East, in 3.5 seconds. What is its acceleration?

We know that: V _{i} = 0 m / s, V _{f} = 21 m / s, t = 3.5 seconds. Therefore we apply the formula:

**a = dv / dt = (v _{f} – v _{i} ) / (t _{f} – t _{i} ), that is, a = 21m / s / 3.5 s = 6 m / s ^{2}**

- A car reduces its speed from 21m / s East to 7m / s East in 3.5.0 seconds. What is its acceleration?

Knowing that V _{i} = 21 m / s, V _{f} = 7 m / s, t = 3.5 seconds, and that a = dv / dt = (v _{f} – v _{i} ) / (t _{f} – t _{i} ), is simple calculate it:

**a = 7m / s – 21m / s / 3.5s = -4m / s ^{2} **, that is, a negative acceleration (deceleration).