Respuesta :
Answer:
Step-by-step explanation:
a) since acceleration can be expressed as a(t) = dv/dt
We use the chosen rule to obtain
dv/dt = dv/dx * dx/dt = v* dv/dx
We can now set v dv/dx = -kv^2
Taking v from both sides
dv/dx = -kv
Dv/v = -kdx
In (v)= kx+ c
V= e^kx+c
From v (o) = 150
V = 150e^kx
When distance x= 2000ft convert to miles gives 0.378788 mi then velocity v = 15 mi/he
15= 150. e ^ 0.378788k
Lm (o.t) = -0.378788k
K= 6.0788mi
C) we return to dv/dt = -kv^2 = dv/v^2 = -k dt
-1/v = -kt +c
From v(o) = 150, we get - 1/150= c and
1/v - 1/150= kt
We know v= 15mi/hr and k = 6.0788
1/15 - 1/150 = 6.0788t
T = 9.8704.10^-3hr.
The velocity of the rocket sled can be given by the function [tex]v = 150e^{-kx}[/tex].
What is the velocity of the sled?
We know that acceleration can be written as,
[tex]a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot \dfrac{dx}{dt}[/tex]
we know velocity can also be written as,
[tex]v=\dfrac{dx}{dt}[/tex]
therefore, the acceleration can be written as,
[tex]a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot \dfrac{dx}{dt}= \dfrac{dv}{dx}\cdot v[/tex]
Substituting the value of a,
[tex]-kv^2 = \dfrac{dv}{dx}\cdot v\\\\\dfrac{-kv^2}{v} = \dfrac{dv}{dx}[/tex]
rearranging the terms,
[tex]-kdx = \dfrac{dv}{v}[/tex]
Integrating both sides,
[tex]\int-kdx = \int \dfrac{dv}{v}[/tex]
[tex]-kx+c= ln\ v[/tex]
taking Antilog,
[tex]v = e^{-kx+c}[/tex]
[tex]v = e^{-kx}e^c[/tex]
[tex]v = Ce^{-kx}[/tex]
given to us the initial speed of the sled was 150 mi/hr.,
[tex]150 = Ce^{-k(0)}[/tex]
[tex]C=150[/tex]
[tex]v = 150e^{-kx}[/tex]
Hence, the velocity of the sled can be given by the function [tex]v = 150e^{-kx}[/tex].
What is the value of k?
It is given to us distance traveled by the sled is 2000 ft at a speed of 15 mi/hr.
[tex]15 = 150e^{-k(2000)}[/tex]
[tex]0.1 = e^{-k(2000)}\\\\ln\ 0.1 = -k(0.378788)\\\\-2.3= -k(0.378788)\\\\k= 6.07199[/tex]
Thus, the value of the constant k is 1.1513 x 10⁻³.
Find how many seconds of braking is required to slow the sled to 15 mi/hr?
We know that acceleration can be written as,
[tex]a =\dfrac{dv}{dt}\\\\-kv^2 =\dfrac{dv}{dt}\\\\-k\ dt = \dfrac{dv}{v^2}\\\\\int -kdt =\int \dfrac{dv}{v^2}\\\\-k\int dt =\int \dfrac{dv}{v^2}\\\\-kt+c =\dfrac{-1}{v}\\\\\dfrac{1}{v}=kt-c[/tex]
We know the initial speed of the sled is 150, at t=0.
[tex]c=-\dfrac{1}{150}[/tex]
Substitute the values we get,
[tex]\dfrac{1}{15}=1.1513 \times 10^{-3}t-\dfrac{1}{150}[/tex]
[tex]t= 9.8704\times 10^{-3}\rm\ hr[/tex]
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