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By the end of this section, you will be able to:
  • Calculate the acceleration vector given the velocity function in unit vector notation.
  • Describe the motion of a particle with a constant acceleration in three dimensions.
  • Use the one-dimensional motion equations along perpendicular axes to solve a problem in two or three dimensions with a constant acceleration.
  • Express the acceleration in unit vector notation.

Instantaneous acceleration

In addition to obtaining the displacement and velocity vectors of an object in motion, we often want to know its acceleration vector    at any point in time along its trajectory. This acceleration vector is the instantaneous acceleration and it can be obtained from the derivative with respect to time of the velocity function, as we have seen in a previous chapter. The only difference in two or three dimensions is that these are now vector quantities. Taking the derivative with respect to time v ( t ) , we find

a ( t ) = lim t 0 v ( t + Δ t ) v ( t ) Δ t = d v ( t ) d t .

The acceleration in terms of components is

a ( t ) = d v x ( t ) d t i ^ + d v y ( t ) d t j ^ + d v z ( t ) d t k ^ .

Also, since the velocity is the derivative of the position function, we can write the acceleration in terms of the second derivative of the position function:

a ( t ) = d 2 x ( t ) d t 2 i ^ + d 2 y ( t ) d t 2 j ^ + d 2 z ( t ) d t 2 k ^ .

Finding an acceleration vector

A particle has a velocity of v ( t ) = 5.0 t i ^ + t 2 j ^ 2.0 t 3 k ^ m/s . (a) What is the acceleration function? (b) What is the acceleration vector at t = 2.0 s? Find its magnitude and direction.

Solution

(a) We take the first derivative with respect to time of the velocity function to find the acceleration. The derivative is taken component by component:

a ( t ) = 5.0 i ^ + 2.0 t j ^ 6.0 t 2 k ^ m/ s 2 .

(b) Evaluating a ( 2.0 s ) = 5.0 i ^ + 4.0 j ^ 24.0 k ^ m/ s 2 gives us the direction in unit vector notation. The magnitude of the acceleration is | a ( 2.0 s ) | = 5.0 2 + 4.0 2 + ( −24.0 ) 2 = 24.8 m/ s 2 .

Significance

In this example we find that acceleration has a time dependence and is changing throughout the motion. Let’s consider a different velocity function for the particle.

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Finding a particle acceleration

A particle has a position function r ( t ) = ( 10 t t 2 ) i ^ + 5 t j ^ + 5 t k ^ m . (a) What is the velocity? (b) What is the acceleration? (c) Describe the motion from t = 0 s.

Strategy

We can gain some insight into the problem by looking at the position function. It is linear in y and z , so we know the acceleration in these directions is zero when we take the second derivative. Also, note that the position in the x direction is zero for t = 0 s and t = 10 s.

Solution

(a) Taking the derivative with respect to time of the position function, we find

v ( t ) = ( 10 2 t ) i ^ + 5 j ^ + 5 k ^ m/s .

The velocity function is linear in time in the x direction and is constant in the y and z directions.

(b) Taking the derivative of the velocity function, we find

a ( t ) = −2 i ^ m/s 2 .

The acceleration vector is a constant in the negative x -direction.

(c) The trajectory of the particle can be seen in [link] . Let’s look in the y and z directions first. The particle’s position increases steadily as a function of time with a constant velocity in these directions. In the x direction, however, the particle follows a path in positive x until t = 5 s, when it reverses direction. We know this from looking at the velocity function, which becomes zero at this time and negative thereafter. We also know this because the acceleration is negative and constant—meaning, the particle is decelerating, or accelerating in the negative direction. The particle’s position reaches 25 m, where it then reverses direction and begins to accelerate in the negative x direction. The position reaches zero at t = 10 s.

Practice Key Terms 1

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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