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ra1l [238]
3 years ago
13

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.
Mathematics
1 answer:
ANTONII [103]3 years ago
3 0

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.

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