C O N C E P T U A L C H E C K P O I N T
A V E R A G E S P E E D
You drive 4.00 mi at 30.0 mi/h and then another 4.00 mi at 50.0 mi/h. Is your average
speed for the 8.00-mi trip (a) greater than 40.0 mi/h, (b) equal to 40.0 mi/h, or (c) less
than 40.0 mi/h?
R E A S O N I N G A N D D I S C U S S I O N
At first glance it might seem that the average speed is definitely 40.0 mi/h. On further reflection, however, it is clear that it takes more time to travel 4.00 mi at 30.0 mi/h than it does to travel 4.00 mi at 50.0 mi/h. Therefore, you will be traveling at the lower speed for a greater period of time, and hence your average speed will be less than 40.0 mi/h—that is, closer to 30.0 mi/h than to 50.0 mi/h.
A N S W E R
(c) The average speed is less than 40.0 mi/h.
To confirm the conclusion of the Conceptual Checkpoint, we simply apply the definition of average speed to find its value for this trip. We already know that the distance traveled is 8.00 mi; what we need now is the elapsed time. On the first 4.00 mi the time is
t1= 4 mi / 30 mi/h = (4/30)h
The time required to cover the second 4.00 mi is
t2= 4 mi/ 50 mi/h = (4/50)h
Therefore, the elapsed time for the entire trip is
t1 + t2 = (4/30)h + (4/50)h = 0.213 h
This gives the following average speed:
average speed = 8 mi / 0.213 h = 37.6 mi/h < 40.0 mi/h
Note that a “guess” will never give a detailed result like 37.6 mi/h; a systematic, step-by-step calculation is required.
In many situations, there is a quantity that is even more useful than the average speed. It is the average velocity, and it is defined as displacement per time.
BOOKS
Basic Electronics book by Pakistani author
Nuclear physics-1 book by Pakistani author
Nuclear physics complete book by Pakistani author
Solid State Physics by M A Wahab 3rd ed,
Mechanics part 2 by indian author
Electricity and magnetism book by indian author
Methods of Mathematical Physics by Syed Jamal Anwar
Solid state physics by RK Puri VK Babber book pdf
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