A 2750-kg test rocket is launched vertically from the launch pad. Its fuel (of n

Question

A 2750-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t)=At+Bt^2 , where A and B are constants and time is measured from the instant the fuel is ignited. At the instant of ignition, the rocket has an upward acceleration of 1.30m/s^2 and 1.80s later an upward velocity of 2.01m/s .

A.) Determine A

B.) Determine B

C.) At 3.10s after fuel ignition, what is the acceleration of the rocket?

D.) At 3.10s after fuel ignition,what thrust force does the burning fuel exert on it, assume no air resistance? Express the thrust in newtons.

E.) What thrust force does the burning fuel exert on it, assume no air resistance? Express the thrust as a multiple of the rocket's weight.

F.) What was the initial thrust due to the fuel?

Explanation / Answer

given that :

mass of the rocket, m = 2750kg

upward velocity, v = 2.01 m/s

upward acceleartion, a = 1.3 m/s2

time taken, t = 1.8 sec

(a) a given equation, v (t) = At + Bt2                      { eq. 1 }

a (t) = v (t) / t

differentiating w.r.t time,

a (t) = A + 2 Bt                                      { eq. 2 }

at t = 0,    a (0) = A + 2 B (0)

1.3 m/s2 = A + 0

A = 1.3 m/s2

(b) using eq.1,   v (t) = At + Bt2  

at t = 1.8,   v (1.8) = (1.3 m/s2) (1.8 sec) + B (1.8 sec)2

(2.01 m/s) = (2.34 m/s) + (3.24 B) (sec)2

- 0.33 = 3.24 B

B = - 0.101 m/s2

(c) At 3.10 sec after fuel ignition, the acceleration of the rocket is given as :

using eq. 2,     a (t) = A + 2 Bt           

at t = 3.1 sec,    a (3.1) = (1.3 m/s2) + 2 (-0.101 m/s2) (3.1 s)

a (3.1) = (1.3) - (0.6262)   m/s2

a (3.1) = 0.67 m/s2

(d) At 3.10s after fuel ignition, thrust force is given as :

F = ma              { eq.3 }

F = T - mg                                { eq.4 }

equating above eq.3 & 4,

ma = T - mg                                              { eq.5 }

where, g = acceleration due to gravity = 9.8 m/s2

inserting the values in eq. 5,

(2750kg) (0.67 m/s2) = T - (2750kg) (9.8 m/s2)

1842.5 = T - 26950

T = 28792.5 N

(e) Expression for the thrust as a multiple of the rocket's weight which is given as :

T / mg                              { eq.6 }

inserting the values in above eq.

(28792.5 N) / (2750kg) (9.8 m/s2)

(28792.5 N) / 26950 N

1.06 (factor of rocket weight)

(f) the initial thrust due to the fuel is given as :

again using eq. 5,           ma = T - mg         

inserting the values in above eq.

(2750kg) (1.3 m/s2) = T - (2750kg) (9.8 m/s2)

3575 = T - 26950

T = 30525 N

Factor, T / mg = (30525 N) / (2750kg) (9.8 m/s2)

(30525 N) / (26950)

1.13 (times the weight of the rocket)

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