Can an Object at a Constant Acceleration Slow Down Then Speed Up Again?

Learning Objectives

By the end of this section, you will be able to:

• Define and distinguish between instantaneous acceleration, average acceleration, and deceleration.
• Calculate acceleration given initial time, initial velocity, concluding time, and final velocity.

Figure 1. A airplane decelerates, or slows down, as information technology comes in for landing in St. Maarten. Its acceleration is reverse in direction to its velocity. (credit: Steve Conry, Flickr)

In everyday conversation, to advance means to speed up. The accelerator in a car can in fact crusade it to speed upwards. The greater the acceleration, the greater the modify in velocity over a given time. The formal definition of acceleration is consistent with these notions, but more inclusive.

Average Acceleration

Average Acceleration is the rate at which velocity changes ,

[latex]\bar{a}=\frac{\Delta v}{\Delta t}=\frac{{v}_{f}-{v}_{0}}{{t}_{f}-{t}_{0}}[/latex]

where [latex]\bar{a}[/latex] is boilerplate dispatch, v is velocity, and t is time. (The bar over the a ways average acceleration.)

Because acceleration is velocity in yard/s divided past time in s, the SI units for dispatch are m/s^two , meters per 2nd squared or meters per 2d per 2d, which literally means by how many meters per second the velocity changes every 2d.

Recall that velocity is a vector—it has both magnitude and management. This means that a change in velocity can be a alter in magnitude (or speed), but it can also exist a change in management . For example, if a automobile turns a corner at abiding speed, it is accelerating because its direction is changing. The quicker you turn, the greater the acceleration. And then there is an acceleration when velocity changes either in magnitude (an increment or subtract in speed) or in direction, or both.

Acceleration as a Vector

Acceleration is a vector in the same direction equally the change in velocity, Δ v . Since velocity is a vector, it tin alter either in magnitude or in management. Acceleration is therefore a alter in either speed or direction, or both.

Keep in heed that although dispatch is in the direction of the change in velocity, it is not always in the direction of motion . When an object slows down, its acceleration is opposite to the management of its move. This is known as deceleration.

Instantaneous Dispatch

Instantaneous acceleration a , or the acceleration at a specific instant in time , is obtained past the same process as discussed for instantaneous velocity in Fourth dimension, Velocity, and Speed—that is, by because an infinitesimally modest interval of fourth dimension. How do we observe instantaneous dispatch using only algebra? The answer is that we choose an average dispatch that is representative of the motion. Figure half-dozen shows graphs of instantaneous dispatch versus time for two very different motions. In Figure six(a), the acceleration varies slightly and the boilerplate over the entire interval is about the same every bit the instantaneous acceleration at any time. In this case, we should treat this move equally if it had a constant acceleration equal to the average (in this instance nearly 1.8 g/due south²). In Figure 6(b), the acceleration varies drastically over time. In such situations it is best to consider smaller time intervals and cull an average dispatch for each. For example, we could consider motility over the time intervals from 0 to 1.0 s and from ane.0 to 3.0 s equally separate motions with accelerations of +iii.0 grand/s² and –2.0 k/s² , respectively.

The next several examples consider the motion of the subway railroad train shown in Figure seven. In (a) the shuttle moves to the right, and in (b) it moves to the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the reasoning that goes into solving bug.

Case 2. Calculating Displacement: A Subway Train

What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of Figure 7?

Strategy

A drawing with a coordinate system is already provided, so we don't demand to make a sketch, but we should analyze information technology to brand certain we sympathize what it is showing. Pay detail attention to the coordinate system. To observe deportation, we employ the equation Δ 10 = x _f− ten ₀ . This is straightforward since the initial and final positions are given.

Solution

one. Identify the knowns. In the figure we meet that x _f= 6.70 km and x ₀= iv.70 km for part (a), and 10 ′_f= 3.75 km and x ′₀= five.25 km for office (b).

2. Solve for deportation in part (a).

[latex]\Delta ten={x}_{f}-{x}_{0}=6.70\text{ km}-4.70\text{ km} = \text{+}2.00\text{ km}[/latex]

3. Solve for displacement in part (b).

[latex]\Delta x′ ={x′}_{f}-{ten′}_{0}=\text{3.75 km}-\text{5.25 km} = -\text{one.50 km}[/latex]

Discussion

The direction of the move in (a) is to the correct and therefore its displacement has a positive sign, whereas move in (b) is to the left and thus has a negative sign.

Example 3. Comparison Distance Traveled with Displacement: A Subway Railroad train

What are the distances traveled for the motions shown in parts (a) and (b) of the subway train in Figure seven?

Strategy

To answer this question, retrieve about the definitions of distance and distance traveled, and how they are related to displacement. Distance between two positions is defined to be the magnitude of displacement, which was found in Case 1. Distance traveled is the total length of the path traveled betwixt the two positions. (Encounter Deportation.) In the case of the subway train shown in Figure seven, the distance traveled is the aforementioned as the altitude between the initial and final positions of the train.

Solution

i. The deportation for role (a) was +2.00 km. Therefore, the distance between the initial and final positions was ii.00 km, and the distance traveled was 2.00 km.

2. The displacement for part (b) was −1.5 km. Therefore, the distance between the initial and terminal positions was 1.50 km, and the altitude traveled was i.50 km.

Word

Distance is a scalar. It has magnitude but no sign to point direction.

Instance 4. Computing Dispatch: A Subway Train Speeding Up

Suppose the train in Figure 7(a) accelerates from rest to 30.0 km/h in the offset twenty.0 south of its motility. What is its average acceleration during that fourth dimension interval?

Strategy

It is worth it at this point to make a simple sketch:

Solution

1. Place the knowns. v ₀= 0 (the trains starts at remainder), 5 _f= 30.0 km/h, and Δ t = xx.0 s.

2. Summate Δ v . Since the train starts from rest, its modify in velocity is [latex]\Delta v\text{=}\text{+}\text{30.0 km/h}[/latex], where the plus sign ways velocity to the right.

3. Plug in known values and solve for the unknown, [latex]\bar{a}[/latex].

[latex]\bar{a}=\frac{\Delta v}{\Delta t}=\frac{+\text{30.0 km/h}}{\text{20}\text{.}0 south}[/latex]

iv. Since the units are mixed (we have both hours and seconds for time), nosotros need to convert everything into SI units of meters and seconds. (See Physical Quantities and Units for more guidance.)

[latex]\bar{a}=\left(\frac{+\text{30 km/h}}{\text{20.0 southward}}\correct)\left(\frac{{\text{10}}^{iii}\text{m}}{\text{1 km}}\correct)\left(\frac{\text{1 h}}{\text{3600 due south}}\right)=0\text{.}{\text{417 m/s}}^{2}[/latex]

Discussion

The plus sign means that dispatch is to the right. This is reasonable because the railroad train starts from rest and ends up with a velocity to the right (also positive). So acceleration is in the same direction as the alter in velocity, equally is always the case.

Example five. Summate Dispatch: A Subway Railroad train Slowing Down

Now suppose that at the end of its trip, the train in Figure 7(a) slows to a stop from a speed of 30.0 km/h in eight.00 s. What is its average acceleration while stopping?

Strategy

Solution

1. Place the knowns. v ₀= 30.0 km/h, 5 _f= 0 km/h (the train is stopped, and then its velocity is 0), and Δ t = 8.00 s.

2. Solve for the change in velocity, Δ v .

Δ v = v _f − five ₀ = 0 − thirty.0 km/h = − 30.0 km/h

3. Plug in the knowns, Δ v and Δ t , and solve for [latex]\bar{a}[/latex].

[latex]\bar{a}=\frac{\Delta five}{\Delta t}=\frac{-\text{30}\text{.}\text{0 km/h}}{8\text{.}\text{00 s}}[/latex]

4. Convert the units to meters and seconds.

[latex]\bar{a}=\frac{\Delta v}{\Delta t}=\left(\frac{-\text{30.0 km/h}}{\text{viii.00 southward}}\correct)\left(\frac{{\text{ten}}^{3}\text{m}}{\text{1 km}}\correct)\left(\frac{\text{1 h}}{\text{3600 southward}}\right)={\text{-1.04 m/due south}}^{2}\text{.}[/latex]

Discussion

The minus sign indicates that dispatch is to the left. This sign is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would oppose the motion. Once more, acceleration is in the same management as the change in velocity, which is negative here. This acceleration can exist called a deceleration because information technology has a direction reverse to the velocity.

The graphs of position, velocity, and acceleration vs. fourth dimension for the trains in Instance 4 and Example 5 are displayed in Figure 10. (We take taken the velocity to remain constant from 20 to 40 southward, after which the train decelerates.)

Example 6. Calculating Average Velocity: The Subway Train

What is the average velocity of the railroad train in role b of Case two, and shown once more beneath, if it takes 5.00 min to brand its trip?

Strategy

Average velocity is displacement divided by time. It volition be negative here, since the train moves to the left and has a negative displacement.

Solution

one. Identify the knowns. ten ′_f= iii.75 km, x ′₀= 5.25 km, Δ t = 5.00 min.

ii. Determine displacement, Δ ten ′. We institute Δ x ′ to be −one.5 km in Example 2.

iii. Solve for average velocity.

[latex]\bar{v}=\frac{\Delta ten′}{\Delta t}=\frac{-\text{1.50 km}}{\text{5.00 min}}[/latex]

4. Convert units.

[latex]\bar{5}=\frac{\Delta ten′}{\Delta t}=\left(\frac{-i\text{.}\text{50 km}}{v\text{.}\text{00 min}}\correct)\left(\frac{\text{60 min}}{i h}\correct)=-\text{eighteen}\text{.0 km/h}[/latex]

Discussion

The negative velocity indicates motion to the left.

Instance seven. Computing Deceleration: The Subway Train

Finally, suppose the train in Effigy 2 slows to a finish from a velocity of 20.0 km/h in 10.0 south. What is its average dispatch?

Strategy

Once once again, let'due south depict a sketch:

As earlier, we must discover the change in velocity and the change in time to calculate average acceleration.

Solution

1. Identify the knowns. v ₀= −twenty km/h, v _f= 0 km/h, Δ t = 10.0 s.

2. Calculate Δ v . The change in velocity here is really positive, since

[latex]\Delta five={v}_{f}-{v}_{0}=0-\left(-\text{20 km/h}\correct)\text{=}\phantom{\rule{0.25}{0ex}}\text{+}\text{20 km/h}[/latex]

3. Solve for [latex]\bar{a}[/latex].

[latex]\bar{a}=\frac{\Delta v}{\Delta t}=\frac{+\text{xx}\text{.0 km/h}}{\text{10}\text{.}0 s}[/latex]

four. Catechumen units.

[latex]\bar{a}=\left(\frac{+\text{20}\text{.}\text{0 km/h}}{\text{ten}\text{.}\text{0 s}}\right)\left(\frac{{\text{10}}^{three}k}{one km}\right)\left(\frac{1 h}{\text{3600 s}}\right)\text{=}\text{+}0\text{.556 m}{\text{/s}}^{2}[/latex]

Give-and-take

The plus sign means that acceleration is to the correct. This is reasonable because the train initially has a negative velocity (to the left) in this problem and a positive acceleration opposes the motion (and so information technology is to the right). Again, acceleration is in the aforementioned direction as the change in velocity, which is positive here. As in Instance v, this dispatch can be called a deceleration since it is in the direction opposite to the velocity.

Sign and Direction

Mayhap the most of import matter to notation about these examples is the signs of the answers. In our chosen coordinate system, plus means the quantity is to the right and minus means information technology is to the left. This is piece of cake to imagine for displacement and velocity. Only information technology is a footling less obvious for acceleration. Most people interpret negative acceleration every bit the slowing of an object. This was non the case in Instance 2, where a positive acceleration slowed a negative velocity. The crucial distinction was that the acceleration was in the opposite direction from the velocity. In fact, a negative acceleration will increase a negative velocity. For example, the train moving to the left in Figure 11 is sped upward past an acceleration to the left. In that case, both v and a are negative. The plus and minus signs give the directions of the accelerations. If dispatch has the same sign equally the change in velocity, the object is speeding up. If acceleration has the contrary sign of the change in velocity, the object is slowing down.

Cheque Your Understanding

An plane lands on a rails traveling eastward. Draw its acceleration.

Solution

If we have east to be positive, then the plane has negative dispatch, as it is accelerating toward the due west. It is also decelerating: its dispatch is contrary in direction to its velocity.

PhET Explorations: Moving Man Simulation

Learn most position, velocity, and acceleration graphs. Move the little human being back and forth with the mouse and plot his move. Fix the position, velocity, or acceleration and let the simulation move the homo for you.

Click to download the simulation. Run using Coffee.

Department Summary

• Acceleration is the rate at which velocity changes. In symbols, average dispatch [latex]\bar{a}[/latex] is

  [latex]\bar{a}=\frac{\Delta v}{\Delta t}=\frac{{v}_{f}-{v}_{0}}{{t}_{f}-{t}_{0}}\text{.}[/latex]

• The SI unit of measurement for acceleration is [latex]{\text{m/s}}^{2}[/latex] .
• Acceleration is a vector, and thus has a both a magnitude and direction.
• Acceleration tin can be caused by either a change in the magnitude or the direction of the velocity.
• Instantaneous acceleration a is the dispatch at a specific instant in time.
• Deceleration is an acceleration with a direction opposite to that of the velocity.

Conceptual Questions

1. Is it possible for speed to exist constant while acceleration is not zero? Requite an example of such a situation.

2. Is it possible for velocity to be abiding while acceleration is not naught? Explain.

3. Give an example in which velocity is zero yet acceleration is non.

4. If a subway railroad train is moving to the left (has a negative velocity) and so comes to a cease, what is the management of its dispatch? Is the acceleration positive or negative?

v. Plus and minus signs are used in one-dimensional motion to indicate management. What is the sign of an acceleration that reduces the magnitude of a negative velocity? Of a positive velocity?

Problems & Exercises

1. A cheetah can advance from rest to a speed of 30.0 yard/south in 7.00 south. What is its acceleration?

ii. Professional person Application.Dr. John Paul Stapp was U.S. Air Forcefulness officer who studied the effects of farthermost deceleration on the human trunk. On Dec 10, 1954, Stapp rode a rocket sled, accelerating from remainder to a height speed of 282 m/s (1015 km/h) in five.00 s, and was brought jarringly back to rest in only 1.40 due south! Calculate his (a) dispatch and (b) deceleration. Express each in multiples of g (9.lxxx k/s²) past taking its ratio to the acceleration of gravity.

3. A commuter backs her car out of her garage with an acceleration of 1.40 m/s².(a) How long does information technology accept her to achieve a speed of two.00 m/s? (b) If she then brakes to a end in 0.800 s, what is her deceleration?

iv. Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 due south (the actual speed and time are classified). What is its average acceleration in m/s² and in multiples of1000 (9.fourscore chiliad/sⁱⁱ).

Glossary

acceleration:

the rate of change in velocity; the alter in velocity over fourth dimension

average acceleration:

the change in velocity divided past the time over which information technology changes

instantaneous acceleration:

dispatch at a specific point in time

deceleration:

acceleration in the direction reverse to velocity; acceleration that results in a decrease in velocity

Selected Solutions to Problems & Exercises

1. 4.29 m/sⁱⁱ

three. (a) ane.43 s (b) -2.l yard/sⁱⁱ

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Source: https://courses.lumenlearning.com/physics/chapter/2-4-acceleration/

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