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Bad White [126]
2 years ago
9

As a glacier melts, the volume V of the ice, measured in cubic kilometers, decreases at a rate modeled by the differential equat

ion dVdt=kV , where t is measured in years. The volume of the glacier is 400km3 at time t=0 . At the moment when the volume of the glacier is 300km3 , the volume is decreasing at the rate of 15km3 per year. What is the volume V in terms of time
Physics
1 answer:
vovangra [49]2 years ago
7 0

Solve the differential equation:

d<em>V</em>/d<em>t</em> = <em>k V</em>   →   1/<em>V</em> d<em>V</em>/d<em>t</em> = <em>k</em>

→   d/d<em>t</em> [ln(<em>V</em>)] = <em>k</em>

→   ln(<em>V</em>) = <em>k t</em> + <em>C</em>

→   <em>V</em> (<em>t</em> )= exp(<em>k t</em> + <em>C </em>) = <em>C</em> exp(<em>k t</em> ) = <em>C e </em>ᵏᵗ

At <em>t</em> = 0, the glacier has volume 400 km³ of ice, so

<em>V</em> (0) = 400   →   <em>C</em> <em>e</em>⁰ = <em>C</em> = 400

Find when the glacier's volume is 300 km³:

<em>V</em> (<em>t</em> ) = 400 <em>e </em>ᵏᵗ = 300   →   <em>e </em>ᵏᵗ = 3/4

→   <em>k t</em> = ln(3/4)

→   <em>t</em> = 1/<em>k</em> ln(3/4)

At this time, the volume is decreasing at a rate of 15 km³/yr, so

<em>V '</em> (<em>t </em>) = <em>C</em> <em>k</em> <em>e </em>ᵏᵗ   →   <em>V '</em> (1/<em>k</em> ln(3/4)) = 400 <em>k</em> exp(<em>k</em> × 1/<em>k</em> ln(3/4)) = -15

→   3/4 <em>k</em> = -3/80

→   <em>k</em> = -1/20

Then the volume <em>V</em> (<em>t</em> ) of the glacier at time <em>t</em> is

<em>V</em> (<em>t</em> ) = 400 exp(-1/20 <em>t </em>)

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Una tractomula se desplaza con rapidez de 69 km/h. Cuando el conductor ve una vaca atravesada enmedio de la carretera, acciona l
Anestetic [448]

Answer:

Los datos que tenemos:

Rapidez: 69km/h

Tiempo que tarda en frenar = 4s.

Distancia inicial entre la tracto-mula y la vaca = 25m

Ok, la ecuación de desaceleración es:

D = (sf - si)/t

sf = velocidad final = 0m/s

si = velocidad inicial = 69km/h

t = tiempo = 4s

D = -69km/h/4s

ok, 1h = 3600s

D = (-69km/s)*1/(4*3600s)  = -0.0048 km/s^2

Entonces la ecuación de aceleración es:

a(t) =  -0.0048 km/s^2

Para la velocidad, integramos sobre el tiempo

v(t) = (-0.0048 km/s^2)*t + v0

donde v0 es la velocidad inicial, en este caso v0 = 69km/3600s = 0.0191km/s  

v(t) =  (-0.0048 km/s^2)*t + 0.0191km/s

Para la posición volvemos a integrar sobre el tiempo, esta vez suponemos la posición inicial igual a cero.

p(t) = (1/2)*(-0.0048 km/s^2)*t^2 + 0.0191m/s*t

Ahora, si p(t=4s) < 25m, esto implica que la tracto-mula no impacto con la vaca.

p(4s) = (1/2)*(-0.0048 km/s^2)*(4s)^2 + 0.0191km/s*4s = 0.038km

y 1km = 1000m

0.038km = 0.038*1000m = 38m

Entonces si, atropello a la vaca.

4 0
3 years ago
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