Instant acceleration what it is, how it is calculated and exercises

The instantaneous acceleration It is the change that the speed experiences per unit of time at each instant of the movement. At the precise moment when the “dragster”Of the image was photographed, it bore an acceleration of 29.4 m / s^two. This means that at that time, its velocity was being increased by 29.4 m / s in the span of 1 s. This is equivalent to 105 km / h in just 1 second.

A dragster competition is easily modeled assuming the racing car is a point object. P that moves in a straight line. On that line, choose an axis oriented with origin OR that we will call the axisOX) or just axis x.

The kinematic variables that define and describe the movement are:

• The position x
• Displacement Δx
• Speed v
• Acceleration to

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

In the case of rectilinear motion there are only two possible directions: positive (+) in the direction of (OX) or negative (-) in the opposite direction of (OX). Therefore, it is possible to dispense with the formal vector notation and use the signs to indicate the sense of the magnitude.

Article index

• 1 How is acceleration calculated?
• 2 Velocity from instantaneous acceleration
• 3 Solved exercises
  • 3.1 Exercise 1
  • 3.2 Exercise 2
  • 3.3 Exercise 3
  • 3.4 Exercise 4
• 4 References

How do you calculate the acceleration?

Suppose that at the instant t the particle has speed v (t) and in the instant t ' its speed is v (t ').

So the change that the speed had in that period of time was Δv = v (t ') - v (t). Hence the acceleration in the time lapse Δt = t '- t , would be given by the quotient:

This quotient is the average acceleration at_m in the period of time Δt between the instants t and t '.

If we wanted to calculate the acceleration just at time t, then t 'should be an insignificantly greater quantity than t. With this Δt, which is the difference between the two, should be almost zero.

Mathematically it is indicated as follows: Δt → 0 and it is obtained:

I) A particle moves along the X axis with constant velocity v₀ = 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 constant velocity is zero.

II) A particle moves on the axis x and its speed changes with time according to the following formula:

v (t) = 2 - 3t

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

The result is interpreted like this: for any instant the acceleration is -3 m / s.

Between instants 0 s and 2/3 s, the speed is positive while the acceleration is negative, that is, in that interval the particle decreases its speed or decelerates.

At the instant 2/3 s, its velocity becomes zero, but as an acceleration of -3 m / s remains, from that instant the velocity reverses (it becomes negative).

In the instants after ⅔ s, the particle accelerates, since each time its speed becomes more negative, that is, its speed (speed modulus) grows.

III) The figure shows a curve that represents the velocity as a function of time, for a particle moving along the X axis. Find the sign of the acceleration at times t₁, t_two and you₃. Also indicate if the particle is accelerating or decelerating.

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

For the instant t₁, the slope is negative, so the acceleration is negative. And since at that instant the velocity is positive, we can affirm that at that instant the particle is decelerating.

For the instant t_two the tangent line to the curve v (t) is horizontal, so its slope is zero. The mobile has zero acceleration, therefore at t_two the particle neither accelerates nor decelerates.

For the instant t₃, the slope of the tangent line to the curve v (t) is positive. With a positive acceleration, the particle actually accelerates, because at that moment the velocity is also positive..

Velocity from instantaneous acceleration

In the previous section, instantaneous acceleration was defined from instantaneous velocity. In other words, if the speed is known at each instant, then it is also possible to know the acceleration at each instant of the movement..

The reverse process is possible. That is, knowing the acceleration for each instant, then the instantaneous velocity can be calculated.

If the operation that allows going from velocity to acceleration is the derivative, the opposite mathematical operation is integration.  

Solved exercises

Exercise 1

The acceleration of a particle moving along the X axis is a (t) = ¼ t^two. Where t is measured in seconds and a in m / s. Determine the acceleration and velocity of the particle at 2 s of motion, knowing that at the initial instant t₀ = 0 was at rest.

Answer

At 2 s the acceleration is 1 m / s^two and the speed 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^two - 2 t, where t is measured in seconds. Determine the acceleration at times: 0s, 1s, 3s.

Answers

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

a (t) = 6t -2

Then a (0) = -2 m / s^two ; a (1) = 4 m / s^two ; a (3) = 16 m / s^two .

Exercise 3

A metal sphere is released from the top of a building. Falling acceleration is the acceleration of gravity that can be approximated by the value 10 m / s2 and pointing downward. Determine the speed of the sphere 3 s after it has been released.

Answer

This problem involves the acceleration of gravity. Taking the vertical direction as positive down, we have that the acceleration of the sphere is:

a (t) = 10 m / s^two 

And the speed will be given by: 

Exercise 4

A metal sphere is shot upward with an initial speed of 30 m / s. The acceleration of motion is the acceleration of gravity that can be approximated by the value 10 m / s^two and pointing down. Determine the speed of the sphere at 2 s and 4 s after it has been shot.

Answer

The vertical direction will be taken as positive upwards. ANDIn this case the acceleration of the movement will be given by

a (t) = -10 m / s^two   

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

After 4 s of having been fired, the velocity will be 30 - 10 ∙ 4 = -10 m / s. Which means that at 4 s the sphere is descending with a speed of 10 m / s.

References

1. Giancoli, D. Physics. Principles with Applications. 6th Edition. Prentice Hall. 25-27.
2. Resnick, R. (1999). Physical. Volume 1. Third edition in Spanish. Mexico. Compañía Editorial Continental S.A. de C.V. 22-27.
3. Serway, R., Jewett, J. (2008). Physics for Science and Engineering. Volume 1. 7th. Edition. Mexico. Cengage Learning Editors. 25-30.