Assume that a long straight track connects the different concourses of a large airport, and that a train is restricted to move back and forth on this track.  This train is an example of an object, which is restricted to move along a line.  Let us call the line the x-axis of our coordinate system and choose some point on the line as our origin.  The object's position vector can then point in the positive or negative x-direction, and the object can move in the positive or negative x-direction.

In one dimension, if the x-component of a vector is positive, the vector is pointing in the positive x-direction, and if the x-component of a vector is negative, the vector is pointing in the negative x-direction.
In one dimension, the sign of the number functions as the direction indicator.

For motion in one dimension we may write v = v_x i, or v = vi, since the velocity has only an x component, or just v, where a positive number is a velocity vector pointing in the +x-direction and a negative number is a velocity vector pointing in the negative -direction.

Assume that at a time t₁ it is at a position x₁ and at a later time t₂ it is at position x₂. 
The displacement vector is ∆x = x₂ - x₁. 
In the time interval ∆t = t₂ - t₁ the object has been displaced by ∆x.
Its average velocity in that time interval ∆t is defined as <v> = ∆x/∆t. 
The average velocity <v> is a vector. 
The sign of <v> tells us if the velocity vector is pointing in the positive or negative x-direction.

The speed of an object is the rate at which it covers distance.  It is always a positive number with units. 
The average speed is the total distance traveled in the time interval ∆t.  It is a scalar, i.e. it has no direction. 
We write

average speed <v> = distance/time = d/∆t.

If you cover a distance of 80 miles in two hours, then your average speed is

<v> = (80 miles)/(2 hours) = 40 miles/hour.

According to the rules of algebra, we can rewrite the formula <v> = d/∆t in two different ways.

• If we want to know how the distance a car, going at 55 miles/hour, travels in 3 hours we write d = <v>∆t = (55 miles/hour)*(3 hours) = 165 miles.
• If we want to know how long it will take this car to cover a distance of 220 miles we write ∆t = d/<v> = (220 miles)/(55 miles/hour) = 4 hours.

The units, such as miles and hours are always carried along in calculations and are treated like ordinary algebraic quantities.

Problem:

A motorist drives north for 35 minutes at 85 km/h and then stops for 15 minutes.  He then continues north, traveling 130 km in 2 hours. 
(a)  What is his total displacement?
(b)  What is his average velocity? 

Solution:

• Reasoning:
  To find the total displacement we add the two given displacements.  To find the average velocity, we divide the total displacement vector by the total time it takes to get from the initial to the final position.
• Details of the calculation:
  (a)  In the first 35 minutes the motorist travels
  d₁ = v₁t₁ = 85 km/h * 35 min*1 h/(60 min) = 49.6 km, north.
  In the next 2 hours he travels 130 km, north.
  The total displacement is 179.6 km. north.
  
  (b)  Let the positive x-direction be north.
  His average velocity is <v> = ∆x/∆t.
  He travels for 170 minutes (including his stop).
  Therefore his average velocity is
  <v> = (179.6 km/(170 min))*(60 min/h) (north) = 63.4 km/h (north).

Often the average velocity is not a very useful quantity.  We want to know the velocity of an object at an instant of time.  We want to know the instantaneous velocity.  The more we reduce the time interval between two successive position measurements, the closer is the average velocity measured for that time interval to the instantaneous velocity.  We define the instantaneous velocity as the limit of ∆x/∆t, as ∆t approaches zero.

The instantaneous speed is the magnitude of the instantaneous velocity.  From now on we will assume that the words velocity and speed stand for instantaneous velocity and instantaneous speed unless explicitly stated otherwise.

External link:  Average vs. Instantaneous Speed

Problem:

Can the magnitude of the average velocity be larger than the average speed?

Solution:

• Reasoning:
  No, for an objects moving along a straight line without reversing its direction, the average speed is the magnitude of the average velocity.  If the object turns around or if it does not move along a straight line, then the average speed is larger than the magnitude of the average velocity.

When the velocity of an object is changing, the object is accelerating.
Acceleration is the rate at which the velocity is changing.
In one dimension the velocity is changing the speed is changing.
The object can speed up, slow down, or come to a stop and turn around.  The key word again is CHANGE.

In one dimension <a> = ∆v/∆t = (v₂ - v₁)/(t₂ - t₁).  The sign of a tells us if the acceleration vector is pointing in the positive or negative x-direction.

Consider a time interval ∆t = 1 s.  Consider the following values for v₁ and v₁.

Note:  Positive acceleration does not always mean increase in speed and negative acceleration does not always mean decrease in speed.

Problem:

You are traveling east at 30 miles per hour.  You see a ball rolling onto the road and you break hard, because you are afraid that a child will come running after the ball.  You come to a stop in 0.8 seconds.  What is the direction of your average velocity in this short time interval?  What is your average acceleration?

Solution:

• Reasoning:
  The direction of your average velocity is east.
  You are traveling east with decreasing speed until you come to a stop.  The direction of your average acceleration is west, opposite the direction of v, since you are slowing down
• Details of the calculation:
  The change in your velocity is ∆v = v₂ - v₁ = 0 - 30 mph east = -30 mph east = 30 mph west.
  Let us convert to SI units.
  30 mph times 1609 m/mile times 1 h/(3600 s) = 13.4 m/s.
  Your average acceleration is ∆v/∆t = (13.4 m/s)/(0.8 s) west = 16.8 m/s² west.

Problem:

A woman backs her car out of her garage with an average acceleration of 1.40 m/s².
(a)  How long does it take her to reach a speed of 2.0 m/s?
(b)  If she then brakes and stops in 0.80 s, what is her average acceleration while braking?

Solution:

• Reasoning:
  Let the direction out of the garage be the positive x- direction.
  The velocity of the car first increases, then decreases again.
  The acceleration of the car is first positive and then negative.
• Details of the calculation:
  (a)  ∆v = <a>*∆t.  Here the initial velocity is zero, ∆v = v_f - v_i = v_f.
  ∆t = ∆v/a = v_f/<a> = (2 m/s)/(1.4 m/s²) = 1.43 m/s.
  (b)  <a> = ∆v/∆t = (v_f - v_i)/∆t.  Here v_f = 0, v_i = 2 m/s.
  <a> = (-2 m/s)/0.80 s = -2.5 m/s.

Embedded Question 3

Consider motion in one dimension.
(a)  When do the velocity and the acceleration vector have the same direction?
(b)  When do the velocity and the acceleration vector have opposite direction?
(c)  Can an object that is momentarily at rest be accelerating?

Discuss this with your fellow students in the discussion forum!

Real, extended objects

Assume that you are riding in a car, which is traveling down the road.  The speedometer is not working, but you would like to determine the average speed of the car.  You note that at time t₁ = 8:31:00 AM the car passes mile marker 93 and at time t₂ = 8:32:05 AM the car passes mile marker 94.  You write down the positions of the car at times t₁ and t₂ as marker 93 and marker 94 respectively.  Is this good enough?

The car is an extended object.  Can we just treat it like a moving point particle?  The answer depends on the problem we are trying to solve.  How accurate do you want your answer to be?  Do you just want to find the average speed of the car to make sure you are not exceeding the speed limit?  Then noting when the car passes the mile markers is probably good enough to determine its speed.  But if you are trying to determine the outcome of a race, it is be very important to note if the front bumper, the center, or the rear bumper of the car passes a marker at a given time.

When we treat extended objects as point particles, we neglect their orientation and their internal motion.  For many problems this is an acceptable approximation, but it is always an approximation.  We then set the position of the center of the object equal to the position of the object.