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

Answer:

(a) $s(44) = 3.52*10^{6}ft$

(b) $v(44) = 10*x^{5} ft/s$

(c) $a(44) = 3636ft/s^{2}$

Step-by-step 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

$speed,v=\frac{ds(t)}{dt}=\frac{d}{dt}(1818t^{2}) \\\\v = 1818*2t^{2-1}= 1818*2t\\ \\v = 3636t$

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

$acceleration, a = \frac{dv}{dt}=\frac{d}{dt}(3636t)\\ \\a = 3636*1= 3636\\\\a = 3636ft/s^{2}$

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