Accounting for viscous drag and fuel-burning, the one-dimensional equation of motion for a rocket- propelled vehicle travelling along a horizontal surface is dv m dt dm - k dt 1 = -- where (in SI units) p = 1.2 is the density of air, A = 1 is the effective frontal area of the vehicle, Cp = 0. 25 is the drag coefficient and k = 3000 is the rocket's thrust coefficient. Furthermore, if the mass of the vehicle after fuelling is 3000 kg and the fuel burn rate is a linear 6 kg s¯', then one may write т %3D 3000 —6t Substitute this expression for m and the values of p, A, Cp and k into the differential equation above to show that dv 18, 000 – 0. 15v² dt 3000 – 6t and solve this numerically using Euler's method in MATLAB with the initial condition v(0)= 0 and the step size h = 5 (include a screenshot),
Accounting for viscous drag and fuel-burning, the one-dimensional equation of motion for a rocket- propelled vehicle travelling along a horizontal surface is dv m dt dm - k dt 1 = -- where (in SI units) p = 1.2 is the density of air, A = 1 is the effective frontal area of the vehicle, Cp = 0. 25 is the drag coefficient and k = 3000 is the rocket's thrust coefficient. Furthermore, if the mass of the vehicle after fuelling is 3000 kg and the fuel burn rate is a linear 6 kg s¯', then one may write т %3D 3000 —6t Substitute this expression for m and the values of p, A, Cp and k into the differential equation above to show that dv 18, 000 – 0. 15v² dt 3000 – 6t and solve this numerically using Euler's method in MATLAB with the initial condition v(0)= 0 and the step size h = 5 (include a screenshot),
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Transcribed Image Text:Accounting for viscous drag and fuel-burning, the one-dimensional equation of motion for a rocket-
propelled vehicle travelling along a horizontal surface is
dv
dm
m
dt
= -PACDU-k
dt
where (in SI units) p = 1. 2 is the density of air, A = 1 is the effective frontal area of the vehicle,
CD = 0. 25 is the drag coefficient and k = 3000 is the rocket's thrust coefficient. Furthermore, if the
mass of the vehicle after fuelling is 3000 kg and the fuel burn rate is a linear 6 kg s, then one may
%3D
S
write
m =
3000 – 6t
Substitute this expression for m and the values of p, A, Cp and k into the differential equation
above to show that
dv
18, 000 – 0. 15v²
%3D
dt
3000 – 6t
and solve this numerically using Euler's method in MATLAB with the initial condition v(0)= 0
and the step size h = 5 (include a screenshot),
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