Average Acceleration and Instantaneous Acceleration

4.4	Average Acceleration and Instantaneous Acceleration

When a particle’s velocity changes from

in a time interval

, its average acceleration

during

(4-15)

If we shrink

to zero about some instant, then in the limit

approaches the instantaneous acceleration (or acceleration)

at that instant; that is,

(4-16)

If the velocity changes in either magnitude or direction (or both), the particle must have an acceleration.

We can write Eq. 4-16 in unit-vector form by substituting Eq. 4-11 for

to obtain

We can rewrite this as

(4-17)

where the scalar components of

are

(4-18)

To find the scalar components of

, we differentiate the scalar components of

Figure 4-7 shows an acceleration vector

and its scalar components for a particle moving in two dimensions. Caution: When an acceleration vector is drawn, as in Fig. 4-7, it does not extend from one position to another. Rather, it shows the direction of acceleration for a particle located at its tail, and its length (representing the acceleration magnitude) can be drawn to any scale.

FIGURE 4-7

The acceleration

of a particle and the scalar components of

Checkpoint 2

Here are four descriptions of the position (in meters) of a puck as it moves in an xy plane:

1.	and

2.	and

3.

4.

Are the x and y acceleration components constant? Is acceleration

constant?

Sample Problem 4-4

For the rabbit in Sample Problems 4-2 and 4-3, find the acceleration

at time



KEY IDEAS

We can find

by taking derivatives of the rabbit’s velocity components.

Calculations: Applying the a_x part of Eq. 4-18 to Eq. 4-13, we find the x component of

to be

Similarly, applying the a_y part of Eq. 4-18 to Eq. 4-14 yields the y component as

We see that the acceleration does not vary with time (it is a constant) because the time variable t does not appear in the expression for either acceleration component. Equation 4-17 then yields


	(Answer)

which is superimposed on the rabbit’s path in Fig. 4-8.

FIGURE 4-8

The acceleration

of the rabbit at

. The rabbit happens to have this same acceleration at all points on its path.

To get the magnitude and angle of

, either we use a vector-capable calculator or we follow Eq. 3-6. For the magnitude we have


	(Answer)

For the angle we have

However, this angle, which is the one displayed on a calculator, indicates that

is directed to the right and downward in Fig. 4-8. Yet, we know from the components that

must be directed to the left and upward. To find the other angle that has the same tangent as −35° but is not displayed on a calculator, we add 180°:


	(Answer)

This is consistent with the components of

. Note that

has the same magnitude and direction throughout the rabbit’s run because the acceleration is constant.

Sample Problem 4-5

A particle with velocity

(in meters per second) at t = 0 undergoes a constant acceleration

of magnitude

at an angle

from the positive direction of the x axis. What is the particle’s velocity



KEY IDEA

Because the acceleration is constant, Eq. 2-11 (

) applies, we must apply it separately for motion parallel to the x axis and motion parallel to the y axis.

Calculations: We find the velocity components

and

from the equations

In these equations,

and

are the x and y components of

, and

and

are the x and y components of

. To find

and

, we resolve

either with a vector-capable calculator or with Eq. 3-5:

When these values are inserted into the equations for

and

, we find that, at time

Thus, at

, we have, after rounding,


	(Answer)

Either using a vector-capable calculator or following Eq. 3-6, we find that the magnitude and angle of

are


	(Answer)

and


	(Answer)

Check: Does 127° appear on your calculator’s display, or does −53° appear? Now sketch the vector

with its components to see which angle is reasonable.

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