Question

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?

Answer 1

Answer:

t = 1.068 s

Step-by-step 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)

$\dfrac{dv}{dt}= - kv$

$\dfrac{dv}{v}= - kdt$

integrating both side

$\int_{800}^{700}\dfrac{dv}{v}= - k\int dt$

$ln(\dfrac{7}{8})=-kt$

$t = -\dfrac{1}{k}ln(\dfrac{7}{8})$..........(1)

since,

$v = v_1e^{-kt}$

$\dfrac{ds}{dt} = 800e^{-kt}$

$\int ds= 800\int_0^t e^{-kt}dt$

$1400= -\dfrac{800}{k}[e^{-kt}-e^0]$

from equation 1

$k= -\dfrac{8}{14}[\dfrac{7}{8}-1]$

$k= \dfrac{1}{8}$

putting value of k in equation (1)

$t = -8ln(\dfrac{7}{8})$

t = 1.068 s