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A "home-made" solid propellant rocket has an initial mass of 9 kg; 6.8 kg of this is fuel. The rocket is directed vertically upward from rest, burns fuel at a constant rate of 0:225 kg=s, and ejects exhaust gas at a speed of 1980 m=s relative to the rocket. Assume that the pressure at the exit is atmospheric and that air resistance may be neglected. Calculate the rocket speed after 20 s and the distance traveled by the rocket in 20 s. Plot the rocket speed and the distance traveled as functions of time.

Respuesta :

jolis1796

Answer:

v = 1176.23 m/s

y = 741192.997 m = 741.19 km

Explanation:

Given

M₀ = 9 Kg  (Initial mass)

me = 0.225 Kg/s   (Rate of fuel consumption)

ve = 1980 m/s    (Exhaust velocity relative to rocket, leaving at atmospheric pressure)

v = ? if t = 20 s

y = ?

We use the equation

v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt     where t ∈ (0, t)

⇒   v = - ve*Ln ((M₀ - me*t)/M₀) - g*t

then we have

v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)

v = 1176.23 m/s

then we apply the formula

y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt

⇒   y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt

⇒   y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀  - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)

For t = 20 s   we have

y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg  - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)

⇒   y = 741192.997 m = 741.19 km

The graphs are shown in the pics.

Ver imagen jolis1796
Ver imagen jolis1796

Following are the solution to the given points:

Given:

[tex]\to v_o =1980\ \frac{m}{s}\\\\[/tex]

[tex]\to a=g= -9.8 \frac{m}{s} \ \text{(Diection downwords )}[/tex]

Solution:

Using formula:

[tex]\to V= V_o +at\\\\[/tex]

[tex]\to v^2-v^2_0= 2ad \\\\[/tex]

For point a:

[tex]\to V= V_o +at\\\\[/tex]

        [tex]=1980+ (-9.8) (20) \\\\[/tex]

        [tex]= 1980-196 \\\\ =1784\ \frac{m}{s}\\\\[/tex]

The velocity at the conclusion of a [tex]20[/tex] second is [tex]\bold{1784\ \frac{m}{s}}\\\\[/tex].

For point b:

Using formula:

[tex]\to v^2-v^2_0= 2ad \\\\[/tex]

[tex]\to d=\frac{v^2_0 -V^2}{2a}\\\\[/tex]

       [tex]= \frac{(1980)^2 - (1784)^2}{2\times 9.8} \\\\ =\frac{3920400-3182656}{19.6}\\\\= \frac{737744}{17.6}\\\\ =37640\ m \\\\ = 37.64\ km\\\\[/tex]

In [tex]20[/tex] seconds, the total distance traveled is [tex]401737 \ \ m (or \ 401.737\ \ km)[/tex].  

For point c:

Please find the attached file.

Ver imagen codiepienagoya
Ver imagen codiepienagoya
Ver imagen codiepienagoya

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