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** = v_{x}
**i**, or **v** = v**i**, 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 t_{1} it is at a position x_{1}
and at a later time t_{2} it is at position x_{2}.

The displacement vector is ∆x =
x_{2 }-
x_{1}.

In the time interval ∆t = t_{2
}- t_{1} 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?

Solution:

(a) In the first 35 minutes the motorist travels

d_{1
}= v_{1}t_{1 }= 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> = (v_{2 }- v_{1})/∆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 v_{1} and v_{1}.

v_{1} |
v_{2} |
a | sign of a |
sign of ∆v (v = speed) |
---|---|---|---|---|

1 m/s | 2 m/s | 1 m/s^{2} |
+ | + |

2 m/s | 1 m/s | -1 m/s^{2} |
- | - |

1 m/s | -1 m/s | -2 m/s^{2} |
- | ∆v = 0 |

-1 m/s | -2 m/s | -1m/s^{2} |
- | + |

-2 m/s | -1 m/s | 1 m/s^{2} |
+ | - |

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/s^{2}.

(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 = v_{f} -
v_{i} = v_{f}.

∆t = ∆v/a = v_{f}/<**a**> = (2 m/s)/(1.4 m/s^{2}) = 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.