# Fuel burn rate

by Simka
Tags: burn, fuel, rate
 P: 1 A rocket, mass m=4000kg, lifts off the ground vertically with acceleration of a=0.2g. The velocity of gass emmiting from it is u=1200m/s (with respect to the rocket). Find engine's reactive force and then fuel burn rate (mu). This is what i managed to do: rocket's speed at any moment t (with respect to the ground) is a*t gass's speed at any moment t (with respect to the ground) is u-a*t Mass of rocket at any time t is m-mu*t Mass of fuel that is burned is mu*t rocket's mass multiplied by it's speed and gass's mass multiplied by its speed are equal, so (m-mu*t)*a*t=mu*t*(u-a*t) and i get that m*a*t=mu*t*u mu=(m*a)/u=6.53 kg/s but... I was given 4 answers (5.9kg/s 9.8kg/s 39.2kg/s 19.8kg/s) and mine is wrong... Any ideas anyone?
$$F_{rocket} = mg + ma = - F_{gas} = - vdm/dt$$