Question

1. The table shows the height, in centimeters, that a weight bouncing from a spring would achieve if there were no friction, for a given number of seconds.

2. How many seconds are required for the pendulum to swing from its position furthest to the right to its position furthest to the left?

3. How many degrees is the painted spot turning in 0.75 s?
Enter your answer in the box.

4. What is the average daily maximum temperature in March?
Round to the nearest tenth of a degree if needed.
Use 3.14 for π .
Enter your answer in the box

5. What is the greatest temperature the substance reached during the experiment?
Round to the nearest tenth of a degree if needed.
Use 3.14 for π .
Enter your answer in the box.

Answer 1

Problem 1

Answer: 6

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At time 1.5s the height is 0 which is the resting position. The next time the height is 0 is when t = 4.5 which is a difference of 3 seconds (4.5 - 1.5 = 3). Double this to get 3*2 = 6

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Problem 2

Answer: 1.25

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Notice how the period of this graph is 2.5 seconds. It takes 2.5 seconds for the up and down pattern to repeat fully again. The easiest way to see this is from x = 0 to x = 2.5 the down and then up pattern does one full cycle, only to repeat again from x = 2.5 to x = 5, and so on.

Half of this cycle is the length of time it takes for the particle to go from the peak to the valley, or vice versa. Half of 2.5 is 1.25 (2.5/2 = 1.25). Why is it half? Because a full cycle takes place from one valley to the next neighboring valley. The same happens with the peaks as well.

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Problem 3

Answer: 180 degrees

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A full cycle takes 1.5 seconds, which consists of the wheel rotating 360 degrees (a full revolution)
Half a cycle is 1.5/2 = 0.75 seconds which is half a revolution 360/2 = 180 degrees

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Problem 4

Answer: 61

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We plug in x = 2 since x = 0 corresponds to january, x = 1 corresponds to february, and so on

f(x) = 12.2*cos(pi*x/6) + 54.9
f(2) = 12.2*cos(pi*2/6) + 54.9
f(2) = 12.2*cos(3.14*2/6) + 54.9
f(2) = 12.2*cos(3.14/3) + 54.9
f(2) = 12.2*cos(1.04666666666667) + 54.9
f(2) = 12.2*0.5004596890082 + 54.9
f(2) = 6.10560820590003 + 54.9
f(2) = 61.0056082059
f(2) = 61

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Problem 5

Answer: 38.9

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The smallest cos(x) can be is -1. It doesn't matter if its cos(x) or cos(2x) or whatever the argument is. Therefore f(x) maxes out at f(x) = -7.5*(-1)+31.4 = 38.9