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.
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 Dt = t2 - t1 the object has been displaced by d. Its average velocity in that time interval Dt is defined as v = d/Dt. The displacement d is a vector, the time interval Dt 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 Dt. The distance is a scalar, i.e. a number with units. Dividing the distance by Dt 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?
| ||
| 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?
|
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.
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 
, where
Dr is the
displacement in the time interval Dt. 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.
Assume a car is moving on a circular track with a 1000m 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/p 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/Ö2 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 |
| Can the magnitude of the average velocity be larger than the average speed?
|
Whenever the instantaneous 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 Dv = v2 - v1 in the time interval Dt = t2 - t1. The object's average acceleration in that time interval Dt is defined as a = Dv/Dt. 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 
, where
Dv is the change in
velocity in the time interval Dt. From now on we will assume
that the word acceleration stand for instantaneous acceleration, unless explicitly stated otherwise.
| 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 drive at a constant speed of 20 miles/h around a city block, checking out the
neighborhood. Are you accelerating?
| ||
| 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?
| ||
| 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?
|
Links:
| Football Physics, Speed, Velocity, Acceleration |
| Acceleration |
| Uniform circular Motion |
Exercise (You can earn up to 5 points extra credit by completing this exercise.)
Links to other Web Materials:
| Lesson 1:
Describing Motion with Words An Introduction to the Language of Kinematics
|
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