# Instant acceleration: what it is, how it is calculated and exercised

The **instantaneous acceleration** is the change experienced by velocity per unit of time at each moment of movement. At the exact moment when the “ *dragster* ” of the image was photographed, there was an acceleration of 29.4 m / s ^{2} . This means that, at that time, its velocity was being increased by 29.4 m / s in the interval of 1 s. That’s equivalent to 105 km/h in just 1 second.

A dragster competition is easily modeled, assuming the race car is a point object *P* that moves in a straight line. In this line, a source axis oriented is chosen with *the* that call axle ( *X* ) or simply axis *x* .

Dragsters are racing cars capable of developing huge accelerations. Source: Pixabay.com

The kinematic variables that define and describe the movement are:

- position
*X* *Δx*offset- speed
*v* - Acceleration
*to*

They are all vector quantities. Therefore, they have a magnitude, a direction and a meaning.

In the case of straight motion, there are only two possible directions: positive (+) in the direction of ( *OX* ) or negative (-) in the opposite direction of ( *OX* ). Therefore, formal vector notation can be dispensed with and signs used to indicate the meaning of magnitude.

**How is acceleration calculated?**

Suppose that at time *t* the particle has velocity *v (t)* and at time *t ‘* its velocity is *v (t’)* .

Therefore, the change that velocity had in that period of time was *Δv **= v (t ‘) – v (t). *Therefore, the acceleration in the time interval *Δ **t = t ‘- t* , would be given by the quotient:

This quotient is the average acceleration in _{m} over the time period Δt between instants tet ‘.

If we wanted to calculate the acceleration just at moment t, then t ‘ should be a quantity that is insignificantly larger than t. This Δ t, which is the difference between the two, must be almost zero.

Mathematically, it is indicated like this: Δ t → 0 and you get:

**I)** A particle moves on the X axis with constant velocity v= 3 m / s. What will be the acceleration of the particle?

The derivative of a constant is zero; therefore, the acceleration of a particle moving with a constant velocity is zero.

**II)** A particle moves on the *x-* axis and its velocity changes with time, according to the following formula:

v (t) = 2 – 3t

Where velocity is measured in m/s and time in s. What will be the acceleration of the particle?

The result is interpreted as follows: *at any time the acceleration is -3 m / s* .

Between 0 and 2/3 s, the velocity is positive while the acceleration is negative, that is, in this interval the particle slows down or slows down.

At 2/3 s, its velocity becomes zero, but since an acceleration of -3 m / s remains, from that moment on the velocity is reversed (becomes negative).

In the following moments, the particle is accelerating, because each time its velocity becomes more negative, that is, its velocity (speed modulus) is increasing.

**III)** The figure shows a curve that represents the velocity as a function of time, for a particle that moves on the X axis. Find the sign of the acceleration at moments t _{1} , t _{2} and t _{3} . Also indicate whether the particle is accelerating or decelerating.

Velocity versus time graph for a particle. The slopes of the lines indicate acceleration at the indicated times. Source: own elaboration.

Acceleration is the derivative of the velocity function, so it is equivalent to the slope of the line tangent to the curve v (t) for a given time t.

At time t _{1} , the slope is negative and the acceleration is negative. And since at this moment the velocity is positive, we can say that at this moment the particle is slowing down.

At time t _{2,} the line tangent to the curve v(t) is horizontal, so its slope is zero. The cell has zero acceleration, so at t _{2} the particle neither accelerates nor decelerates.

For instant t _{3} , the slope of the tangent line to the curve v (t) is positive. With positive acceleration, the particle is actually accelerating, because at that time the velocity is also positive.

**Instantaneous Acceleration Velocity**

In the previous section, instantaneous acceleration was defined from instantaneous velocity. That is, if the velocity is known at all times, it is also possible to know the acceleration at each moment of the movement.

The reverse process is possible. In other words, the known acceleration for each instant and the instantaneous velocity can be calculated.

If the operation that makes it possible to go from velocity to acceleration is the derivative, the opposite mathematical operation is integration.

**solved exercises**

**Exercise 1**

The acceleration of a particle moving on the X axis is a (t) = ¼ t ^{2} . Where t is measured in seconds already in m / s. Determine the acceleration and velocity of the particle at 2 s of motion, knowing that at the initial moment t = 0 was at rest.

**Response**

At 2 s, the acceleration is 1 m / s ^{2} and the velocity for instant t will be given by:

**Exercise 2**

An object moves along the X axis with a velocity in m / s, given by:

v (t) = 3 t ^{2} – 2 t, where t is measured in seconds. Determine the acceleration in moments: 0s, 1s, 3s.

**Answers**

Taking the derivative of v (t) with respect to t, the acceleration is obtained at any time:

a(t) = 6t -2

Then at (0) = -2 m / s ^{2} ; one (1) = 4 m / s ^{2} ; a (3) = 16 m / s ^{2} .

**Exercise 3**

A metal sphere is released from the top of a building. The acceleration of fall is the acceleration of gravity that can be approximated by the value 10 m / s 2 and pointing downwards. Determine the speed of the sphere 3 s after it is released.

**Response**

The acceleration of gravity intervenes in this problem. Taking the *downward* vertical direction as positive, the acceleration of the sphere is:

a (t) = 10 m / s ^{2}

And the speed will be given by:

**Exercise 4**

A metal sphere shoots upward with an initial velocity of 30 m/sec. Motion acceleration is the acceleration of gravity that can be approximated by the value 10 m / s ^{2} and pointing downwards. Determine the ball’s velocity in 2 s and 4 s after it is fired.

**Response**

The *upward* vertical direction will be considered positive *. **And* n In this case, the acceleration of motion is given by

a (t) = -10 m / s ^{2}

The speed as a function of time will be given by:

After 4 s of being fired, the velocity will be 30 – 10 ∙ 4 = -10 m / s. Which means that at 4 s the sphere descends quickly with 10 m / s.