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Calculate acceleration: a subway train slowing down

Now suppose that at the end of its trip, the train in [link] (a) slows to a stop from a speed of 30.0 km/h in 8.00 s. What is its average acceleration while stopping?

Strategy

A velocity vector arrow pointing toward the right with initial velocity of thirty point zero kilometers per hour and final velocity of 0. An acceleration vector arrow pointing toward the left, labeled a equals question mark.

In this case, the train is decelerating and its acceleration is negative because it is toward the left. As in the previous example, we must find the change in velocity and the change in time and then solve for acceleration.

Solution

1. Identify the knowns. v 0 = 30 .0 km/h , v f = 0 km/h (the train is stopped, so its velocity is 0), and Δ t = 8.00 s .

2. Solve for the change in velocity, Δ v size 12{Δv} {} .

Δ v = v f − v 0 = 0 − 30 . 0 km/h = − 30 .0 km/h size 12{Δv=v rSub { size 8{f} } - v rSub { size 8{0} } =0 - "30" "." "0 km/h"= - "30" "." "0 km/h"} {}

3. Plug in the knowns, Δ v size 12{Δv} {} and Δ t , and solve for a - .

a - = Δ v Δ t = − 30 . 0 km/h 8 . 00 s size 12{ { bar {a}}= { {Δv} over {Δt} } = { { - "30" "." "0 km/h"} over {8 "." "00 s"} } } {}

4. Convert the units to meters and seconds.

a - = Δ v Δ t = − 30.0 km/h 8.00 s 10 3 m 1 km 1 h 3600 s = −1.04 m/s 2 . size 12{ { bar {a}}= { {Δv} over {Δt} } = left ( { { - "30" "." "0 km/h"} over {8 "." "00 s"} } right ) left ( { {"10" rSup { size 8{3} } " m"} over {"1 km"} } right ) left ( { {"1 h"} over {"3600 s"} } right )= - 1 "." "04 m/s" rSup { size 8{2} } "." } {}

Discussion

The minus sign indicates that acceleration 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. Again, acceleration is in the same direction as the change in velocity, which is negative here. This acceleration can be called a deceleration because it has a direction opposite to the velocity.

The graphs of position, velocity, and acceleration vs. time for the trains in [link] and [link] are displayed in [link] . (We have taken the velocity to remain constant from 20 to 40 s, after which the train decelerates.)

Three graphs. The first is a line graph of position in meters versus time in seconds. The line begins at the origin and has a concave up shape from time equals zero to time equals twenty seconds. It is straight with a positive slope from twenty seconds to forty seconds. It is then convex up from forty to fifty seconds. The second graph is a line graph of velocity in meters per second versus time in seconds. The line is straight with a positive slope beginning at the origin from 0 to twenty seconds. It is flat from twenty to forty seconds. From forty to fifty seconds the line is straight with a negative slope back down to a velocity of 0. The third graph is a line graph of acceleration in meters per second per second versus time in seconds. The line is flat with a positive constant acceleration from zero to twenty seconds. The line then drops to an acceleration of 0 from twenty to forty seconds. The line drops again to a negative acceleration from forty to fifty seconds.
(a) Position of the train over time. Notice that the train’s position changes slowly at the beginning of the journey, then more and more quickly as it picks up speed. Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the velocity remains constant, the position changes at a constant rate. (b) Velocity of the train over time. The train’s velocity increases as it accelerates at the beginning of the journey. It remains the same in the middle of the journey (where there is no acceleration). It decreases as the train decelerates at the end of the journey. (c) The acceleration of the train over time. The train has positive acceleration as it speeds up at the beginning of the journey. It has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down at the end of the journey.

Calculating average velocity: the subway train

What is the average velocity of the train in part b of [link] , and shown again below, if it takes 5.00 min to make its trip?

The train moves toward the left, from an initial position of 5 point 25 kilometers to a final position of 3 point 75 kilometers.

Strategy

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

Solution

1. Identify the knowns. x ′ f = 3 .75 km , x ′ 0 = 5.25 km , Δ t = 5.00 min .

2. Determine displacement, Δ x ′ . We found Δ x ′ to be − 1.5 km in [link] .

3. Solve for average velocity.

v - = Δ x ′ Δ t = − 1.50 km 5.00 min size 12{ { bar {v}}= { {Δ { {x}} sup { ' }} over {Δt} } = { { - 1 "." "50 km"} over {5 "." "00 min"} } } {}

4. Convert units.

v - = Δ x ′ Δ t = − 1 . 50 km 5 . 00 min 60 min 1 h = − 18 .0 km/h size 12{ { bar {v}}= { {Δx'} over {Δt} } = left ( { { - 1 "." "50"`"km"} over {5 "." "00"`"min"} } right ) left ( { {"60"`"min"} over {1`h} } right )= - "18" "." 0`"km/h"} {}

Discussion

The negative velocity indicates motion to the left.

Calculating deceleration: the subway train

Finally, suppose the train in [link] slows to a stop from a velocity of 20.0 km/h in 10.0 s. What is its average acceleration?

Strategy

Once again, let’s draw a sketch:

A velocity vector arrow pointing to the left with initial velocity of negative twenty point 0 kilometers per hour and a final velocity of 0. An acceleration vector arrow pointing toward the right, labeled a equals question mark.

As before, we must find the change in velocity and the change in time to calculate average acceleration.

Solution

1. Identify the knowns. v 0 = − 20 km/h , v f = 0 km/h , Δ t = 10 . 0 s .

2. Calculate Δ v size 12{Δv} {} . The change in velocity here is actually positive, since

Δ v = v f − v 0 = 0 − − 20 km/h = + 20 km/h . size 12{Δv=v rSub { size 8{f} } - v rSub { size 8{0} } =0 - left ( - "20 km/h" right )"=+""20 km/h"} {}

3. Solve for a - size 12{ { bar {a}}} {} .

a - = Δ v Δ t = + 20 .0 km/h 10 . 0 s

4. Convert units.

a - = + 20 . 0 km/h 10 . 0 s 10 3 m 1 km 1 h 3600 s = + 0 .556 m /s 2

Discussion

The plus sign means that acceleration is to the right. 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 it is to the right). Again, acceleration is in the same direction as the change in velocity, which is positive here. As in [link] , this acceleration can be called a deceleration since it is in the direction opposite to the velocity.

Sign and direction

Perhaps the most important thing to note 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 it is to the left. This is easy to imagine for displacement and velocity. But it is a little less obvious for acceleration. Most people interpret negative acceleration as the slowing of an object. This was not the case in [link] , 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 [link] is sped up by an acceleration to the left. In that case, both v size 12{v} {} and a size 12{a} {} are negative. The plus and minus signs give the directions of the accelerations. If acceleration has the same sign as the velocity, the object is speeding up. If acceleration has the opposite sign as the velocity, the object is slowing down.

An airplane lands on a runway traveling east. Describe its acceleration.

If we take east to be positive, then the airplane has negative acceleration, as it is accelerating toward the west. It is also decelerating: its acceleration is opposite in direction to its velocity.

Phet explorations: moving man simulation

Learn about position, velocity, and acceleration graphs. Move the little man back and forth with the mouse and plot his motion. Set the position, velocity, or acceleration and let the simulation move the man for you.

Moving Man

Section summary

  • Acceleration is the rate at which velocity changes. In symbols, average acceleration     a - size 12{ { bar {a}}} {} is
    a - = Δ v Δ t = v f − v 0 t f − t 0 . size 12{ { bar {a}}= { {Δv} over {Δt} } = { {v rSub { size 8{f} } - v rSub { size 8{0} } } over {t rSub { size 8{f} } - t rSub { size 8{0} } } } "." } {}
  • The SI unit for acceleration is m/s 2 size 12{"m/s" rSup { size 8{2} } } {} .
  • Acceleration is a vector, and thus has a both a magnitude and direction.
  • Acceleration can be caused by either a change in the magnitude or the direction of the velocity.
  • Instantaneous acceleration a size 12{a} {} is the acceleration at a specific instant in time.
  • Deceleration is an acceleration with a direction opposite to that of the velocity.

Conceptual questions

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

Is it possible for velocity to be constant while acceleration is not zero? Explain.

Give an example in which velocity is zero yet acceleration is not.

If a subway train is moving to the left (has a negative velocity) and then comes to a stop, what is the direction of its acceleration? Is the acceleration positive or negative?

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

Problems&Exercises

A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?

4 . 29 m/s 2 size 12{4 "." "29"`"m/s" rSup { size 8{2} } } {}

Professional Application

Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his (a) acceleration and (b) deceleration. Express each in multiples of g ( 9 . 80 m /s 2 ) by taking its ratio to the acceleration of gravity.

A commuter backs her car out of her garage with an acceleration of 1 . 40 m/s 2 size 12{1 "." "40 m/s" rSup { size 8{2} } } {} . (a) How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her deceleration?

(a) 1 . 43 s size 12{1 "." "43"`s} {}

(b) − 2 . 50 m/s 2 size 12{ - 2 "." "50"`"m/s" rSup { size 8{2} } } {}

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in m/s 2 size 12{"m/s" rSup { size 8{2} } } {} and in multiples of g ( 9 . 80 m /s 2 ) ?

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Source:  OpenStax, College physics arranged for cpslo phys141. OpenStax CNX. Dec 23, 2014 Download for free at http://legacy.cnx.org/content/col11718/1.4
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