Instruction 1-1 Motion, Distance and Time Relationships | Speed and Velocity |
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| Motion, Distance and Time Relationships | ||||||
| CCSTD Science Grade 8 8.1.a., b., c. Math CA GR6 AF 2.3 | ||||||
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The earth and other planets travel around the sun. A bus carries you from home to school. Water flows in the river, and snow falls in the winter. All these events have one common thing: in each event one object moves from one place to another. In other words, through each of these events, an object changes its position. Such an action is called motion.
Basic Factors of a Motion:
Displacement, Distance,
and Elapsed Time
The linear distance between the original position and the final position of an object is called the displacement of the object in motion. For example, in the figure above, the displacement of the person is equal to the length AD, which is different (shorter) than the distance he traveled. If a motion is on a straight line, we can measure the path of the motion on the x-axis and denote the starting and the ending points of the motion by the points x1 and x2, respectively. Then, denoting the distance covered by an object by ∆x, we can set the formula below for the distance. [The notation ∆x (delta) does not mean that and x are multiplied; rather it represents the magnitude of change (∆ ) in the variable x. In physics, we usually use the symbol ∆ to indicate the magnitude of increase or decrease in a quantity.]
The motion of an object takes time to occur. That is, moving from one place to another happens in the course of time. Such a time interval is called the elapsed time.
For example, assume that your school bus picks you up at 7:20 AM and arrives at school at 7:48 AM. Then this motion, that is your movement from home to school, takes place between 7:20 and 7:48 AM. The difference between these times indicates the elapsed time of your motion from home to school. Elapsed Time = 7:48 AM ─ 7:20 AM = 28 minutes We usually denote the starting time of a motion by t1 and ending time by t2, and the elapse time by ∆t. Then,
Types of Motions
Solution. The distance covered by the object is MA
+ AB + BN. Distance MN = 12 ft. + 23 ft. + 9 ft. = 44 ft. The displacement of the object is equal to the distance between the original position M and the final position N. This length is the same as MN which is 44 ft. Thus, in this case, the distance and the displacement are the same. Practice 2. In order to move from the town A to the town D, a driver must drive through the routes AB, BC, and CD. If a person moves from A to D, what are the distance and the displacement in this movement?
Solution. The distance traveled by moving from A to
D is equal to AB + BC + CD. Replacing the names
of these routes with the lengths we find For experiments related to the concept of motion, see the following web pages: http://www.glencoe.com/sec/science/internet_lab/olc.php?olcChapter=801 http://members.tripod.com/qldscienceteachers/junior/expt/physics_motion.html http://physics.about.com/od/mechanicsexp/a/weightofair.htm http://sprott.physics.wisc.edu/demobook/chapter1.htm
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Speed and Velocity (top) |