If a construction worker drops a wrench from the top of a building 450 feet high the wrench will be s(t) = −16t 2 + 450 feet above the ground after t seconds. Determine the velocity of the wrench 2 seconds after it was was dropped. Answer in units of ft/sec.

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Chenk13 Chenk13 · Answer 1 · 2019-09-06T09:53:20+02:00

Answer:

-64ft/s

Explanation:

Given:

[tex]s(t)=-16t^2+450[/tex]

Unknown:

[tex]\frac{ds}{dt}=v(t)=-32t[/tex]

For t = 2:

[tex]v(2)=-32\times 2=-64[/tex]

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facundo3141592 facundo3141592 · Answer 2 · 2021-09-24T18:28:30+02:00

We know that the rate of change of the position of a given object is equal to the velocity of that object.

We will find that the velocity of the wrench after 2 seconds is -64ft/s

Now we know that the height (vertical position) of the wrench is:

s(t) = -16*t^2 + 450

Then to get the velocity equation we need to differentiate this, remember the general rule:

[tex]f(x) = x^n\\\\\frac{df(x)}{dx} = n*x^{n-1}[/tex]

Then we will get that the velocity equation is:

v(t) = 2*(-16)*t + 0

v(t) = -32*t

And because the position is in ft, this velocity will be in ft per second.

Then the velocity of the wrench 2 seconds after it was dropped is given by:

v(2) = -32*2 = -64

Then the velocity of the wrench after 2 seconds is -64ft/s

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