Question

Use the position function s(t) = –16t2 + 800, which gives the height (in feet) of an object that has fallen for t seconds from a height of 800 feet. The velocity at time t = a seconds is given by the following.If a construction worker drops a wrench from a height of 800 feet, how fast will the wrench be falling after 3 seconds?

Answer 1

To calculate the velocity of the falling wrench after 3 seconds, the first derivative of the position function s(t) is taken to get v(t) = -32t. By substituting t with 3, the velocity at three seconds is -96 feet per second.

Step-by-step explanation:

To find how fast the wrench will be falling after 3 seconds, we need to calculate the first derivative of the position function s(t) to get the velocity function v(t). The position function given is s(t) = −16t² + 800. Differentiating this with respect to t gives v(t) = d/dt (-16t² + 800) = -32t.

To find the velocity after 3 seconds, we substitute t = 3 into the velocity equation, which gives v(3) = -32(3) = -96 feet per second.

Therefore, the wrench will be falling at a velocity of -96 feet per second after 3 seconds.

Answer 2

96 ft/s

Step-by-step explanation:

We can start by deriving the equation of the velocity, which is the derivative of the position equation:

`v(t) = (ds)/(dt) = (-16t^2)' + 800' = -32t`

After 3 seconds, the wrench would achieve a speed of

v(3) = -32t = -32*3 = -96 ft/s

So it's falling at the rate of 96 ft per second after 3 s