Question 4 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 1 dm m at -żPAC,v² – k- dt where (in Sl units) p = 1.2 is the density of air, A = 1 is the effective frontal area of the vehicle, C, = 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-1, then one may write m = 3000 – 6t Substitute this expression for m and the values of p, A, C, and k into the differential equation above to show that dv_ 18,000 – 0.15v² dt 3000 – 6t
Question 4 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 1 dm m at -żPAC,v² – k- dt where (in Sl units) p = 1.2 is the density of air, A = 1 is the effective frontal area of the vehicle, C, = 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-1, then one may write m = 3000 – 6t Substitute this expression for m and the values of p, A, C, and k into the differential equation above to show that dv_ 18,000 – 0.15v² dt 3000 – 6t
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Question 4
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
dt
1
m at
-PAC,v² – k'
where (in Sl units) p = 1.2 is the density of air, A = 1 is the effective frontal area of the vehicle, C, =
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-1, then one may write
m = 3000 – 6t
Substitute this expression for m and the values of p, A, C, and k into the differential equation
above to show that
dv 18,000 – 0.15v²
dt
3000 – 6t
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