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An object falls freely in a straight line and experiences air resistance proportional to its? speed; this means its acceleration is ?a(t)=??kv(t), where k is a positive constant and v is the? object's velocity. The speed of the object decreases from 800 ft/s to 700ft/s over a distance of 1400 ft. Approximate the time required for this deceleration to occur?

User Bcat
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1 Answer

4 votes

Answer:

t = 1.068 s

Explanation:

given,

a(t) =- k v(t)

speed of the object decreases from 800 ft/s to 700 ft/s

distance = 1400 ft

time for deceleration = ?

a(t) =- k v(t)


(dv)/(dt)= - kv


(dv)/(v)= - kdt

integrating both side


\int_(800)^(700)(dv)/(v)= - k\int dt


ln((7)/(8))=-kt


t = -(1)/(k)ln((7)/(8))..........(1)

since,


v = v_1e^(-kt)


(ds)/(dt) = 800e^(-kt)


\int ds= 800\int_0^t e^(-kt)dt


1400= -(800)/(k)[e^(-kt)-e^0]

from equation 1


k= -(8)/(14)[(7)/(8)-1]


k= (1)/(8)

putting value of k in equation (1)


t = -8ln((7)/(8))

t = 1.068 s

User Eugene Pawlik
by
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