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antiseptic1488
7 days ago
9

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 ab

ove 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

Mathematics
1 answer:
Inessa [3.9K]7 days ago
3 0

Answer:

Keep in mind:

Speed equals distance divided by time.

Next, we can calculate the average speed for different intervals.

Let's define a model in which the average speed at a specific time t = t0 can be calculated as:

AS(t0) = (y(b) - y(a))/(b - a)

Here, b represents the subsequent t0 value, and a is the earlier one. The mid-point in this segment is t0.

Therefore:

If t0 is set at 100 seconds:

AS(100s) = (400ft - 0ft)/(200s - 0s) = 2ft/s.

If t0 is at 200 seconds:

AS(200s) = (1360ft - 50ft)/(300s - 100s) = 6.55 ft/s.

If t0 is at 300 seconds:

AS(300s) = (3200ft - 400ft)/(400s - 200s) = 14ft/s.

If t0 is at 400 seconds:

AS(400s) = (6250s - 1360s)/(500s - 300s) = 24.45 ft/s.

This implies for one of the options, t = 400s has the highest speed.

This makes sense, as while the timing differences remain uniform at 100 seconds, the variances in the height values are increasing.

Thus, we can determine that the rocket is ascending with greater acceleration, leading to a higher average speed as time progresses.

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