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3 years ago

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Knowing that v = –8 m/s when t = 0 and v = 8 m/s when t = 2 s, determine the constant k. (Round the final answer to the nearest

whole number. You must provide an answer before moving on to the next part.) k = m/s4

1 answer:

docker41 [41]3 years ago

5 0

Answer:

a)We know that acceleration a=dv/dt

So dv/dt=kt^2

dv=kt^2dt

Integrating we get

v(t)=kt^3/3+C

Puttin t=0

-8=C

Putting t=2

8=8k/3-8

k=48/8

k=6

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2 years ago

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Luba_88 [7]

Answer:

16 minutes

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This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.

The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."

Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...

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<em>Additional comment</em>

The oil levels at that time will be 4 ft.

You can write two equations for height:

y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)

y = 8 -0.25x . . . . . . height in feet after x minutes (tank B)

These will be equal when ...

y = y

12 -0.5x = 8 -0.25x

4 = 0.25x . . . . . . . . . . add 0.5x -8

16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height

The graph shows when the tanks will have equal heights and when they will be drained.

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