Question

When an object is droppednbsp on a certain earth dash like planet comma on a certain earth-like planet, the distance it falls in t​ seconds, assuming that air resistance is​ negligible, is given by ​s(t)equals=1818t2 where​ s(t) is in feet. Suppose that a​ medic's reflex hammer is dropped from a hovering helicopter. Find​(a) how far the hammer falls in 44 ​sec, ​(b) how fast the hammer is traveling 44 sec after being​ dropped, and ​(c) the​ hammer's acceleration after it has been falling for 44 sec.

Answer 1

(a)
`s(44) = 3.52*10^(6)ft`

(b)
`v(44) = 10*x^(5) ft/s`

(c)
`a(44) = 3636ft/s^(2)`

Explanation:

`s(t) = 1818t^(2)`

(a) when t = 44sec

`s(44) = 1818(44^(2)) = 1818(1936)\\ \\s(44) = 3519648ft`

s(44) ≅
`3520000=3.52*10^(6)ft`

(b) How fast the hammer is traveling i.e. the speed of the hammer.

To find speed, we differentiate the distance s(t) with respect to time, t

`speed,v=(ds(t))/(dt)=(d)/(dt)(1818t^(2)) \\\\v = 1818*2t^(2-1)= 1818*2t\\ \\v = 3636t`

at t = 44sec,

`v = 3636*44=159984ft/s`

`v` ≅
`160000ft/s = 1.6*10^(5)ft/s`

(c) The hammer's acceleration can be obtained by differentiating the speed v(t) with respect to time, t

`acceleration, a = (dv)/(dt)=(d)/(dt)(3636t)\\ \\a = 3636*1= 3636\\\\a = 3636ft/s^(2)`

There's no need to substitute t = 44sec because the acceleration is independent of time.