1. If the position of an object changes from 34 cm at clock time 4.6 seconds to 87 cm at clock time 5.3
seconds, then during this interval what is the average rate of change of its position with respect to clock time?
53cm= change in distance (87-34)
5.3-4.6= change in time
Avg rate of change= change in quantity a/change in time
= 75.7cm/sec
2. If the velocity of an object changes from 12 cm/second at clock time 6.9 seconds to 20 cm/s at clock time
15.3 seconds, then what is the rate of change of its velocity with respect to clock time?
Change in distance=8cm
<h3>right idea but it is not a change in position you need, and the change in position for this interval is not 8 cm</h3>
Change in time= 8.4 sec
Velocity= 8cm/8.4sec=
.95cm/sec

<h3>average rate of change of its velocity with respect to clock time = (change in velocity) / (change in clock time), by definition of average rate.

so we need change in velocity, not change in position

The velocity changes from 12 cm/s to 20 cm/s, which is a change in velocity of 8 cm/s

You have the correct change in clock time.

Thus

average rate of change of its velocity with respect to clock time = (change in velocity) / (change in clock time) = (8 cm/s) / (8.4 s) = .95 cm/s^2.</h3>
3. What is your best estimate of the average velocity of the object in #2, for the given time interval?
.95cm/sec

<h3>velocity changes from 12 cm/s to 20 cm/s. It is reasonable to conjecture that the average velocity is halfway between the initial and final velocities. So without additional information, an average velocity of 16 cm/s is the most reasonable conjecture.

It should be obvious that 16 cm/s is 'right in the middle' of the given initial and final velocity, but we sometimes need to calculate the average of two velocities. The calculation is simple enough--we simply average the initial and final velocities:

ave of initial and final velocity = (initial velocity + final velocity) / 2 = (vf + v0) / 2, so that in this case we get
ave of initial and final velocity = (12 cm/s + 20 cm/s) / 2 = (32 cm/s) / 2 = 16 cm/s.</h3>

4. If the average rate of change of position with respect to clock time during a certain interval is 24 meters /
second, and if the interval lasts for 5 seconds, then what quantity can you determine by applying the definition of an
average rate of change? Find this quantity and explain in detail how you found it.
24*5= 120 meters/5 sec
If it takes only one second to go 24 meters then if we waited 5 seconds we would multiply the distance by 5 to go 120
meters in 5 seconds.
5. What is wrong with saying the average velocity = position / clock time?
Because average velocity would be finding the average or changed clock time and position.

<h3>average rate of change of position with respect to clock time = (change in position) / (change in clock time)

This is very different from position / clock time, which would require only on position and clock time.

To find change in position and change in clock time we need to know two positions and two clock times.</h3>

Text-related questions (to be submitted after completing text assignment)
1. What is the percent uncertainty in a measured time interval of 3.4 seconds, given that the timing mechanism has an
uncertainty of +- .1 second? What is the percent uncertainty in a time interval of .87 seconds, measured using the same
mechanism? When using this mechanism, how does the percent uncertainty in measuring a time interval depend on the
duration of that interval?
.1/3.4*100= 2.9%
.1/.87 *100=11.5%

<h3>note that for equal uncertainties, the shorter the interval the greater the percent difference</h3>

2. What is the uncertainty in the following reported measurements, and what is the percent uncertainty in each?
• 5.8 centimeters +or - .2= 3.4%
• 2350 kilometers +or- 5=0.2%
• 350. seconds +-5= 1.4
• 3.14 +- .06= 1.9%
• 3.1416 +or- .0004=0.013%
3. What is the uncertainty in the area of a rectangle, based on reported length 23.7 cm and width 18.34 cm? 434.7cm^2
uncertainty +-.01

<h3>A reported length of 23.7 cm means that the length rounds off to 23.7 cm. So the length could be anything between 23.65 cm and 23.75 cm. We could write this as 23.7 cm +- .05 cm.

.05 cm is about 0.2% of 23.7 cm, rounded to 1 significant figure..

Similarly 18.34 cm can be written 18.34 cm +- .005 cm.

.005 cm is about 0.03% of 18.34 cm.

When multiplying these two quantities the percent uncertainty of the product will be the sum of the percent uncertainties of the two quantities, so the percent uncertainty is about 0.2% + 0.03% = 0.23%, which still rounds to 0.2%.


If the uncertainty in measurement is regarded as one unit in the last significant figure, which is the convention used in your text, then the numbers are

23.7 cm +- .1 cm, a percent uncertainty of about 0.4% and
18.34 cm +- .01 cm, a percent uncertainty of about .06%

so that the uncertainty in the product is about 0.46%, rounding to 0.5%.</h3>

<h3>A reported length of 23.7 cm means that the length rounds off to 23.7 cm. So the length could be anything between 23.65 cm and 23.75 cm.

Similarly the width could be anywhere between 18.335 cm and 18.345 cm.

If we calculate the area based on the minimum possible values of the length and width, then on the reported values, and finally on the maximum possible values we will obtain areas of , 433.7 cm^2, 434.7 cm^2 and 435.6 cm^2.

The minimum and maximum possible areas could be written as 434.7 cm^2 +- .1 cm.

However 23.7 cm is given to only three significant figures</h3>

4. What is the approximate uncertainty in the area of a circle, based on a reported radius of 2.8 * 10^4 cm?
+-.01

5. What is your height in meters, and your ideal mass in kilograms? How much uncertainty do you think there is in each,
and why?

1.524 meters= 5’5 feet about +- .005uncertainty because I may not have been exact in measuring and through the
conversion I left of figures.
140lb= 63.5 kg about .005 for the same reason