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Minchanka [31]
4 years ago
11

A rocket car is developed to break the land speed record along a salt flat in Utah. However, the safety of the driver must be co

nsidered, so the acceleration of the car must not exceed 5g (or five times the acceleration of gravity) during the test. Using the latest materials and technology, the total mass of the car (including the fuel) is 6000 kilograms, and the mass of the fuel is one-third of the total mass of the car (i.e., 2000 kilograms). The car is moved to the starting line (and left at rest), at which time the rocket is ignited. The rocket fuel is expelled at a constant speed of 900 meters per second relative to the car, and is burned at a constant rate until used up, which takes only 15 seconds. Ignore all effects of friction in this problem.
Find the final acceleration afinal of the car as the rocket is just about to use up its fuel supply. Express your answer to two significant figures.
Physics
1 answer:
Naddika [18.5K]4 years ago
3 0

Answer:

a = 45 m/s/s

Explanation:

As we know that total mass of the rocket is

M = 6000 kg

total mass of the fuel is given as

m = 2000 kg

all the fuel is burnt in 15 s

so rate of the fuel burning is given as

\frac{dm}{dt} = \frac{2000}{15}

\frac{dm}{dt} = 133.33 kg/s

now the thrust force on the rocket is given as

F_{th} = v\frac{dm}{dt}

(6000 - 133.33 t) a = (900 + v)(133.33)

so we have

\frac{dv}{900 + v} = (133.33)\frac{dt}{6000 - 133.33 t}

so we have

ln(\frac{900 + v}{900}) = - ln(\frac{6000 - 133.3 t}{6000})

1 + \frac{v}{900} = \frac{6000}{6000 - 133.3 t}

now acceleration is rate of change in velocity

\frac{1}{900}\frac{dv}{dt} = \frac{133.3\times 6000}{(6000 - 133.3t)^2}

so acceleration at t = 15 s

a = 45 m/s^2

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