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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

A rocket leaves the surface of Earth at time t0 and travels straight up from the surface The height in feet of the rocket above the surface of Earth is given by class=

Respuesta :

facundo3141592

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.

MrRoyal

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:

https://brainly.com/question/359790

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