A rocket leaves the surface of Earth at time t=0 and travels straight up from the surface. The height, in feet, of the rocket above the surface of Earth is given by y(t), where t is measured in seconds for 0≤t≤600. Values of y(t) for selected values of t are given in the table above. Of the following values of t, at which value would the speed of the rocket most likely be greatest based on the data in the table?
(1) t=100
(2) t=200
(3)t=300
(4) t=400
Answers
Answer:
Remember that:
Speed = distance/time.
Then we can calculate the average speed in any segment,
Let's make a model where the average speed at t = t0 can be calculated as:
AS(t0) = (y(b) - y(a))/(b - a)
Where b is the next value of t0, and a is the previous value of t0. This is because t0 is the middle point in this segment.
Then:
if t0 = 100s
AS(100s) = (400ft - 0ft)/(200s - 0s) = 2ft/s
if t0 = 200s
AS(200s) = (1360ft - 50ft)/(300s - 100s) = 6.55 ft/s
if t0 = 300s
AS(300s) = (3200ft - 400ft)/(400s - 200s) = 14ft/s
if t0 = 400s
AS(400s) = (6250s - 1360s)/(500s - 300s) = 24.45 ft/s
So for the given options, t = 400s is the one where the velocity seems to be the biggest.
And this has a lot of sense, because while the distance between the values of time is constant (is always 100 seconds) we can see that the difference between consecutive values of y(t) is increasing.
Then we can conclude that the rocket is accelerating upwards, then as larger is the value of t, bigger will be the average velocity at that point.
Speed is simply the rate of change of distance over time.
The greatest speed is at [tex]t = 400[/tex]
To calculate the time of the greatest speed, we simply calculate the slope between each interval.
The slope (m) of a line is:
[tex]m = \frac{y_2 - y_1}{t_2 - t_1}[/tex]
When [tex]t = 100[/tex]
[tex](t_1,y_1) = (0,0)[/tex]
[tex](t_2,y_2) = (200,400)[/tex]
So, we have:
[tex]m = \frac{400-0}{200 -0}[/tex]
[tex]m= \frac{400}{200}[/tex]
[tex]m = 2[/tex]
When [tex]t = 200[/tex]
[tex](t_1,y_1) = (100,50)[/tex]
[tex](t_2,y_2) = (300,1360)[/tex]
So, we have:
[tex]m = \frac{1360-50}{300 -100}[/tex]
[tex]m = \frac{1310}{200}[/tex]
[tex]m = 6.55[/tex]
When [tex]t = 300[/tex]
[tex](t_1,y_1) = (200,400)[/tex]
[tex](t_2,y_2) = (400,3200)[/tex]
So, we have:
[tex]m = \frac{3200-400}{400 -200}[/tex]
[tex]m = \frac{2800}{200}[/tex]
[tex]m = 14[/tex]
When [tex]t = 400[/tex]
[tex](t_1,y_1) = (300,1360)[/tex]
[tex](t_2,y_2) = (500,6250)[/tex]
So, we have:
[tex]m = \frac{6250-1360}{500-300}[/tex]
[tex]m = \frac{4890}{200}[/tex]
[tex]m = 24.45[/tex]
From the above computation, the greatest speed (i.e. slope) is
[tex]m = 24.45[/tex]
The corresponding time at [tex]m = 24.45[/tex] is [tex]t = 400[/tex]
Hence, the greatest speed is at [tex]t = 400[/tex]
Read more about speed at:
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Dear student,
I request you to make you question more clear.
thank you,
________ ___________________________________
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Juan ran the lemonade stand for 3 more days
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We need more information to be able to answer your question.
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Answer:
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Complete the remaining columns in the chart by calculating the missing values for each
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In a group of 53 children, 22 had a dog, 18 had a cat, and 6 had both a
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Answer:
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Step-by-step explanation:
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Step-by-step explanation:
An airplane flies at an altitude of 14,690 ft while a submarine is submerged at a depth of 521 feet below the surface of the ocean.. What is the difference between these two elevations? Write an algebraic expression for the given applied problem. Simplified the algebraic expression. In the context of the applied problem, explain what your simplified algebraic expression represents.
Answers
Answer:
[tex]d_{T} = d_{A}-d_{S}[/tex]. The algebraic equation represents the distance of the airplane with respecto to the submarine. The distance between the two elevations is 15211 feet.
Step-by-step explanation:
Let be [tex]d[/tex] the distance between a vehicle and sea level, [tex]d > 0[/tex] when vehicle is above sea level, otherwise is below sea level. The difference between airplane and submarine ([tex]d_{T}[/tex]), measured in feet, is obtained by the following expression, that is:
[tex]d_{T} = d_{A}-d_{S}[/tex]
Where:
[tex]d_{A}[/tex] - Distance of airplane with respect to sea level, measured in meters.
[tex]d_{S}[/tex] - Distance of submarine with respect to sea level, measured in meters.
If we know that [tex]d_{A} = 14690\,ft[/tex] and [tex]d_{S} = -521\,ft[/tex], then:
[tex]d_{T} = 14690\,ft - (-521\,ft)[/tex]
[tex]d_{T} = 15211\,ft.[/tex]
The algebraic equation represents the distance of the airplane with respecto to the submarine. The distance between the two elevations is 15211 feet.
The algebraic equation represents the distance of the airplane with respect to the submarine is,
[tex]\rm The \ difference \ between \ both \ elevation = x_a- x_s[/tex]
And the distance between the two elevations is 15,211 feet.
Given that,
An airplane flies at an altitude of 14,690 ft,
While a submarine is submerged at a depth of 521 feet below the surface of the ocean.
We have to determine,
What is the difference between these two elevations?
According to the question,
Let, x be the distance between a vehicle and sea level when the vehicle is above sea level, otherwise is below sea level.
The difference between an airplane [tex]\rm x_a[/tex] and [tex]\rm x_s[/tex] a submarine, measured in feet, is obtained by the following expression, that is:
[tex]\rm The \ difference \ between \ both \ elevation = x_a- x_s[/tex]
Where [tex]\rm x_a[/tex] represents the distance of airplane with respect to sea levels.
And [tex]\rm x_s[/tex] represent the distance of the submarine with respect to sea levels.
Submerged at a depth of 521 feet below the surface of the ocean,
[tex]\rm x_s = -521 \ feet[/tex]
Therefore,
The difference between an airplane [tex]\rm x_a[/tex] and [tex]\rm x_s[/tex] a submarine is,
[tex]\rm The \ difference \ between \ both \ elevation = x_a- x_s\\\\ The \ difference \ between \ both \ elevation = 14,690-(-521)\\\\\rm The \ difference \ between \ both \ elevation = 15,211 \ feet[/tex]
Hence, The difference between these two elevations is 15,211 feet.
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M
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Factors that influence a hypothesis test with the t statistic Suppose you conduct a hypothesis test about a population mean when the standard deviation is unknown calculating a t statistic and determining whether to reject the null hypothesis. Assume that your t statistic positive value and that you will reject the null hypothesis when t is very large. For each of the potential the components of your study design or the results listed below, determine the change to the t statistic, select "Increases, " "Decreases, " or "Stays the same" from the dropdown menu in the right-hand column.
Answers
Please find attachment for full question Answer:
1. Stays same
2. Decrease
3. Stays same
4. Decrease
Explanation:
Please find attachment
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Answers
Answer:
See Explanation
Step-by-step explanation:
Let the temperatures of Miami and Bangor be M and B degrees respectively.
According to the given information:
M = 3B + 22... (required equation)
Plug M = 82
82 = 3B + 22
82 - 22 = 3B
60 = 3B
B = 60/3
B = 20 degrees
Thus, the temperature in Bangor is 20 degrees.
The required temperature in Bangor is 52 degrees and the required equation is 82 = 3B + - 22.
Given that,
The temperature in Miami, Florida is 22 degrees warmer than three times the temperature in Bangor, Maine.
The temperature in Miami is 82 degrees.
An equation to determine the temperature in Bangor is to be determined.
In mathematics, it deals with numbers of operations according to the statements. There are four major arithmetic operators, addition, subtraction, multiplication and division,
Here,
Let the temperature in Bangor be B
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Thus, the required temperature in Bangor is 52 degrees.
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Step-by-step explanation:
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On Friday, the temperature was 82°F. The temperature changed by –2°F on Saturday, and then it changed by 5°F on Sunday. What was the temperature on Sunday? How did the temperature change? This is an easy question but a little bit harder to work out! This was a question on a recent quiz I had and I wanted to see if you guys could figure it out?!
Answers
Answer:
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Friday: 82 F
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