Question

A spacecraft is moving in gravity-free space along a straight path when its pilot decides to accelerate forward. He turns on the thrusters, and burned fuel is ejected at a constant rate of 3.0 × 102 kg/s, at a speed (relative to the rocket) of 2.0 × 102 m/s. The initial mass of the spacecraft and its unburned fuel is 3.0 × 104 kg, the initial fuel mass is 104 kg, and the thrusters are on for 30 s. (a) What is the spacecrafts acceleration as a function of time? (b) What are the spacecraft’s 52 4.5 Exercises accelerations at t = 0, 15, 30, and 35 s? (c) What is the thrust (the force applied to the rocket by the ejected fuel) on the spacecraft?

Answer 1

Derivation of acceleration formula in a general case is given in the image:

And because It is a gravity free space, ie. g=0

Thus acceleration =

a) Putting the values acceleration as a time function

acceleration =   -Ans

b)

at t=0 acceleration =   -Ans

at t=15 acceleration = -Ans

at t=30 acceleration = -Ans

After t=30s the thrusters are turned off

thus at t=35 accleration = -Ans

c) Thrust by rocket is given by ejected burned fuel and is equal to right hand side of eq(1)

Ie Thrust = -Ans

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