Speed, velocity, and acceleration

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.

Time

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 = tlater - tearlier 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.


Velocity and Speed

When an object moves through space, its position changes.  Assume that at a time t1 it is at a position P1 and at a later time t2 it is at position P2.  The displacement vector is d = P2 - P1.  In the time interval ∆t = t2 - t1 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.

Problem:

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.  P2- P1 = 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.

Problem:

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:
P2 - P1 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 |<v>| of the average velocity <v>.

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. 

Problem:

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-->0r/∆t = dr/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.)

Example:

carAssume 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 P1, moving east.
At t = 100 s, it is again at point P1, moving east.  Its velocity averaged over 100 s is 0.  
At t = 50 s, the car is at point P2, moving west.  Its displacement vector for the first 50 seconds is d50 = 1000 m/π south = 318 m south.  Its velocity averaged over the first 50 seconds is v = d50/(50 s) south = 6.36 m/s south.
At t = 25 s the car is at point P3. Its displacement vector for the first 25 s is d25 = 318 m/√2 southeast = 225 m southeast. Its velocity averaged over the first 25 s  is v = d25/(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

Question:

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.


Acceleration

Whenever the velocity of an object is changing, then the object is accelerating.  Assume that at a time t1 the object has velocity v1, and at a later time t2 it has velocity v2.  The change in velocity is ∆v = v2 - v1 in the time interval ∆t = t2 - t1.  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-->0a/∆t = da/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.

Question:

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.

Question:

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.

Question:

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 = 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.

Links: AccelerationUniform circular Motion