Question

If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it eject in the first second to have an initial acceleration of 27.0 m/s2 .

Answer 1

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of  27.0 m/s2  is

      $\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

Explanation:

From the question we are told that

   The mass of the rocket is m = 6300 kg

   The velocity at gas is being ejected is  u =  2000 m/s

    The initial acceleration desired is $a = 27.0 \ m/s$

   The time taken for  the gas to be ejected is  t = 1 s

Generally this desired acceleration is mathematically represented as

        $a = \frac{u * \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}$

Here $\frac{\Delta m}{\Delta t }$  is the rate at which gas is being ejected with respect to time

Substituting values

      $27 = \frac{2000 * \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}$

=>   $170100 -27* \frac{\Delta m}{\Delta t} = 2000 * \frac{\Delta m}{\Delta t}$

=>   $170100 = 2027 * \frac{\Delta m}{\Delta t}$

=>   $\frac{\Delta m}{\Delta t} = \frac{170100}{2027}$

=>   $\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$