At time t=0 a 2150 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. This force obeys the

Question

At time t=0 a 2150 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. This force obeys the equation Fx=At2, where t is time, and has a magnitude of 888.93 N when t=1.25 s.

a) What impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired?
b) By how much does the rockets velocity change during this interval?
c) Find the SI value of the constant A, including its units.

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Edana Edana 3 days 2021-07-22T03:27:27+00:00 1 Answers 0 views 0

Answers ( )

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    2021-07-22T03:28:55+00:00

    Explanation:

    Given that,

    Mass of the rocket, m = 2150 kg

    At time t=0 a rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. The force is given by equation :

    F=At^2

    Here F = 888.93 N when t = 1.25 s

    (c) We can find the value of A first as :

    F=At^2\\\\A=\dfrac{F}{t^2}\\\\A=\dfrac{888.93}{(1.25)^2}\\\\A=568.91\ N/s^2

    The value of A is 568.91\ N/s^2.

    (a) Let J is the impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired. It is given in terms of force as :

    J=\int\limits {F{\cdot} dt}

    Limits will be from 2 s to 2+ 4 = 6 s

    It implies :

    J=\int\limits^6_2 {At^2{\cdot} dt}\\\\J=A\int\limits^6_2 {t^2{\cdot} dt}\\\\J=A\dfrac{t^3}{3}|_2^6\\\\J=568.91\times \dfrac{1}{3}\times (6^3-2^3)\\\\J=39444.42\ Ns

    (b) Impulse is also equal to the change in momentum as :

    J=m\Delta v\\\\\Delta v=\dfrac{J}{m}\\\\\Delta v=\dfrac{39444.42}{2150}\\\\\Delta v=18.34\ m/s

    Hence, this is the required solution.

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