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 t_{1 }= 8:31:00 AM the car passes mile
marker 93 and at time t_{2 }= 8:32:05 AM the car passes mile marker 94.
You write
down the positions of the car at times t_{1} and t_{2} 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.

We measure point in time t by using a clock. The SI unit of time is the
second. The clock reading in seconds depends on the choice of origin.
Clocks in different time zones have different readings and the reading of a stop
watch depends on when it was started. A time interval ∆t = t_{later}
- t_{earlier} is defined as the difference between a later time and an
earlier time. It is always a non-negative number (positive or zero).
A interval ∆t is a scalar, a number with units.

When an object moves through space, its position changes. Assume that at a time t_{1}
it is at a position **P**_{1} and at a later time t_{2} it is at
position **P**_{2}. The displacement vector is
**d **= **P**_{2
}- **P**_{1}.
In the time interval ∆t = t_{2 }- t_{1} the object
has been displaced by **d**. Its **average velocity** in
that time interval ∆t is defined as <**v**>** **=
**d**/∆t.
The displacement **d** is a vector, the time interval ∆t is
a scalar. Dividing a vector by a scalar yields another vector. The average
velocity <**v**>
is a vector. The **average speed** is the distance traveled in the time
interval ∆t divided by ∆t. The distance is a scalar, i.e. a number
with units. Dividing the distance by ∆t yields
another number with units. The average speed is therefore
a scalar.

In a time interval of 5 minutes, a runner runs once around a one-mile track.
What is his
average velocity? What is his average speed?

Solution:

After 5 minute the runner returns to his starting position.
**P**_{2}- **P**_{1 }= 0.
The displacement is zero, so his average velocity is zero.

The
**average speed** is the distance traveled in the time
interval ∆t. This distance is one mile.
The average speed
therefore is (1 mile)/(5 minutes) = (12 miles)/(60 minutes) = 12 miles/h.

Note: Speed is a scalar, velocity is a vector.
The average speed is in general not equal
to the average velocity.

A sprinter runs north in a straight line and covers a distance of 100 m in 12 s.
What is her average velocity? What is her average speed?

Solution:**P**_{2 }- **P**_{1} is 100 m north. The average velocity of the
sprinter is <**v**>** **= (100 m)/(12 s) north = 8.33 m/s north. Her average speed is 8.33 m/s.

Note: For objects moving along a straight line without reversing their direction, the
average speed is the magnitude |<

Summary

The **average speed** is the distance covered divided by
the time it took to cover this distance. If a
person walks 1 km west, then turns around and walks 1 km east, the distance this
person covers is 2 km. If it takes 20 minutes to cover this distance then the
average speed is 2 km/ 20 minutes = 2000 m/ (20*60 s) = 1.67 m/s.

The **average velocity **is a vector. It is the
displacement vector pointing from the initial position to the final position,
divided by the time. In the above example the
initial and the final position are the same. The displacement vector is zero.
So the average velocity is zero.

Link: Speed and velocity (Please explore!)

Since the magnitude and direction of the displacement vector depend on the reference frame in which the coordinate system is anchored and at rest, the velocity of and object depends on the reference frame with respect to which it is measured.

A car has moved forward a distance of 6 m, while a child has moved forward
from the back seat to the front seat a distance of 1 m in a time interval of 2
s. Find the average velocity of the child relative to the car and relative
to the road.

Solution:

Using the car as a reference frame and anchoring the coordinate system in
the car, the displacement of the child is **d**
= (1 m)** i**.

Its velocity is <**v**> = (1 m)/(2 s)
**i** = (0.5 m/s)** i**.

Using the road as a reference frame and anchoring the coordinate system on
the road, the displacement of the child is **d**
= (6 m)** i** + (1 m)** i** = (7 m)** i**.

Its velocity is <**v**> = (7 m)/(2 s)
**i** = (3.5 m/s)** i**.

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

**v** = lim_{∆t-->0}∆**r**/∆t = d**r**/dt,

where
∆**r** is the displacement in the time interval ∆t. 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.

A short review of derivatives (Please review if you are taking calculus concurrently.)

Assume a car is moving on a circular track with a 1000 m circumference. It moves
with constant speed v = 10 m/s. At t = 0 it is at point P_{1}, moving east.

At t = 100 s, it is again at point P_{1}, moving east. Its velocity averaged over
100 s is 0.

At t = 50 s, the car is at point P_{2}, moving west. Its displacement vector for
the first 50 seconds is **d**_{50} = 1000 m/π south = 318 m south. Its velocity
averaged over the first 50 seconds is **v** = d_{50}/(50 s) south = 6.36 m/s south.

At t = 25 s the car is at point P_{3}. Its displacement vector for the first 25 s
is **d**_{25} = 318 m/√2 southeast = 225 m southeast. Its velocity averaged over the
first 25 s is **v** = **d**_{25}/(25 s) southeast = 9 m/s southeast.

The shorter the time interval between successive position measurements, the
closer is the average velocity to the instantaneous velocity, which is 10 m/s
tangential to the circle.

Link: Average vs. Instantaneous Speed

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

Answer:

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.

Since the magnitude and direction of the displacement vector depend on the reference frame in which the coordinate system is anchored and at rest, the velocity of and object depends on the reference frame with respect to which it is measured.

Whenever the velocity of an object is changing, then the object is
accelerating. Assume that at a time t_{1} the object has velocity
**v**_{1},
and at a later time t_{2} it has velocity **v**_{2}. The change in
velocity is ∆**v **= **v**_{2 }-
**v**_{1}
in the time interval ∆t = t_{2 }- t_{1}.
The object's **average acceleration** in that time interval
∆t is defined as <**a**>** **= ∆**v**/∆t.
The average acceleration
<**a**> is a vector.
It is the velocity vector at the final time, minus the velocity vector at
the initial time, divided by the time interval.

We define the **instantaneous acceleration** as

**a** = lim_{∆t-->0}∆**a**/∆t = d**a**/dt,

where ∆**v** is the change in velocity
in the time interval ∆t. From now on we will assume that the word acceleration
stand for instantaneous acceleration, unless explicitly stated otherwise.

Link: Acceleration

Note: Whenever your velocity is
**CHANGING**, you are accelerating.
You are accelerating when you
**CHANGE** your speed, **CHANGE** your direction of travel, or both.
The keyword is **CHANGE**.

If your velocity is **NOT** changing, then, no matter haw fast
you are moving, you are **NOT** accelerating.

You are driving at a constant speed of 20 miles/h around a city block, checking out the
neighborhood. Are you accelerating?

Answer:

Yes, the direction of the velocity vector is changing all the time.

As your turn your car suddenly to the left, loose objects on the dashboard start sliding
to the right. Are these objects accelerating to the right?

Answer:

No, the car is accelerating, since the direction of its velocity is changing.
The objects
on the dashboard keep on moving with the same velocity until they hit something.

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?

Answer:

The direction of your average velocity is east. You are traveling east with
decreasing speed until you come to a stop. The change in your velocity is ∆**v** =
**v**_{2 }- **v**_{1} = 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/s^{2} west.

Links: Acceleration, Uniform circular Motion