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.
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?
(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.
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?
(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.