<< Chapter < Page Chapter >> Page >

The next several examples consider the motion of the subway train shown in [link] . 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 problems.

In part (a), a subway train moves from left to right from an initial position of x equals 4 point 7 kilometers to a final position of x equals 6 point 7 kilometers, with a displacement of 2 point 0 kilometers. In part (b), the train moves toward the left, from an initial position of 5 point 25 kilometers to a final position of 3 point 75 kilometers.
One-dimensional motion of a subway train considered in [link] , [link] , [link] , and [link] . Here we have chosen the x -axis so that + means to the right and means to the left for displacements, velocities, and accelerations. (a) The subway train moves to the right from x 0 to x f . Its displacement Δ x is +2.0 km. (b) The train moves to the left from x 0 to x f size 12{ { {x}} sup { ' } rSub { size 8{f} } } {} . Its displacement Δ x size 12{Δx'} {} is 1 .5 km . (Note that the prime symbol (′) is used simply to distinguish between displacement in the two different situations. The distances of travel and the size of the cars are on different scales to fit everything into the diagram.)

Calculating acceleration: a subway train speeding up

Suppose the train in [link] (a) accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval?

Strategy

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

A point represents the initial velocity of 0 kilometers per second. Below the point is a velocity vector arrow pointing to the right, representing the final velocity of thirty point zero kilometers per hour. Below the velocity vector is an acceleration vector arrow labeled a equals question mark.

This problem involves three steps. First we must determine the change in velocity, then we must determine the change in time, and finally we use these values to calculate the acceleration.

Solution

1. Identify the knowns. v 0 = 0 size 12{v rSub { size 8{0} } =0} {} (the trains starts at rest), v f = 30 . 0 km/h size 12{v rSub { size 8{f} } ="30" "." "0 km/h"} {} , and Δ t = 20 . 0 s size 12{Δt="20" "." "0 s"} {} .

2. Calculate Δ v size 12{Δv} {} . Since the train starts from rest, its change in velocity is Δ v = + 30.0 km/h size 12{Δv"=+""30" "." 0`"km/h"} {} , where the plus sign means velocity to the right.

3. Plug in known values and solve for the unknown, a - size 12{ { bar {a}}} {} .

a - = Δ v Δ t = + 30.0 km/h 20 . 0 s size 12{ { bar {a}}= { {Δv} over {Δt} } = { {+"30" "." 0`"km/h"} over {"20" "." 0`s} } } {}

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

a - = + 30 km/h 20.0 s 10 3 m 1 km 1 h 3600 s = 0 . 417 m/s 2 size 12{ { bar {a}}= left ( { {+"30 km/h"} over {"20" "." "0 s"} } right ) left ( { {"10" rSup { size 8{3} } " m"} over {"1 km"} } right ) left ( { {"1 h"} over {"3600 s"} } right )=0 "." "417 m/s" rSup { size 8{2} } } {}

Discussion

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

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.

Practice Key Terms 4

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Kinematics. OpenStax CNX. Sep 11, 2015 Download for free at https://legacy.cnx.org/content/col11878/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Kinematics' conversation and receive update notifications?

Ask