## Motion in one dimension

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 can function as the direction indicator.
For motion in one dimension we may write v = vx 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 instantaneous velocity is v = lim∆t-->0∆x/∆t = dx/dt.

The speed of an object is the rate at which it covers distance.  It is always a positive number with units.

#### 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:
(a)  In the first 35 minutes the motorist travels
d1 = v1t1 = 85 km/h * 35 min*1 h/(60 min) = 49.6 km.
In the next 2 hours he travels 130 km.

(b)  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).

In one dimension the average acceleration of an object in the time interval ∆t is given by
<a> = (v2 - v1)/∆t = ∆v/∆t,
and the instantaneous acceleration is given by
a = lim∆t-->0∆v/∆t = dv/dt.

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

v1 v2 a sign of a sign of ∆v
(v = speed)
1 m/s 2 m/s 1 m/s2 + +
2 m/s 1 m/s -1 m/s2 - -
1 m/s -1 m/s -2 m/s2 - ∆v = 0
-1 m/s -2 m/s -1m/s2 - +
-2 m/s -1 m/s 1 m/s2 + -

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

#### 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 accceleration?

Solution:
(a)  ∆v = <a>*∆t.  Here the initial velocity is zero, ∆v = vf - vi = vf.
∆t = ∆v/a = vf/<a> = (2 m/s)/(1.4 m/s2) = 1.43 m/s.

(b)  <a> = ∆v/∆t = (vf - vi)/∆t.  Here vf = 0, vi = 2 m/s.
<a> = (-2 m/s)/0.80 s = -2.5 m/s.