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Doss [256]
3 years ago
5

A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket m

otor is ignited and the rocket burns fuel at a rate of 95 kg/s. The exit speed of the exhaust gas relative to the rocket is 2900 m/s. Neglecting drag and friction forces, determine the acceleration and the velocity of the car at time t = 10 s. Plot the acceleration and velocity from time t0 to t = 10 s.

Physics
1 answer:
slamgirl [31]3 years ago
3 0

Answer: Acceleration of the car at time = 10 sec is 108 m/s^{2} and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

M\frac{dv}{dt} = u\frac{dM}{dt}

\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt

       = u\int_{M_{o}}^{M_{f}} \frac{dM}{M}

v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}

v_{o} = 0

As, v_{f} = u ln (\frac{M_{f}}{M_{o}})

u = -2900 m/s

M_{f} = M_{o} - m \times t_{f}

           = 2500 kg + 1000 kg - 95 kg \times t_{f}s

           = (3500 - 95t_{f})s

v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s

Also, we know that

     a = \frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}

        = \frac{u}{3500 - 95 t} \times (-95) m/s^{2}

        = \frac{95 \times 2900}{3500 - 95t} m/s^{2}

At t = 10 sec,

v_{f} = 918.34 m/s

and,   a = 108 m/s^{2}

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