[DiffEq] First order Modeling Applications

[metiscus]

YATP - Yet Another Trainlike Problem

Sailboats A and B each have mass 60kg and cross the starting line at the same time of a race. Each has an initial velocity of 2m/s.
Obviously from this, m1 = m2 == 60kg and vo1 = v02 == 2 m/s.
The wind applies a constant force of 650N to each boat and the water resistance is proportional to the velocity of the boat.
Boat A:
proportionality constants are b1 = 80 Nsec/meter before planing when the velocity is less than 5m/s and b2=60Nsec/m when velocity is aboce 5m/s.

Boat b:
proportionality constants are b1 = 100 Nsec/meter before planing when the velocity is less than 6m/s and b2=50Nsec/m when velocity is aboce 6m/s.

The race is 500m long, which sailboat will be leading at the end of the race?


I assume that the method of solution will be to complete the equations of motion for both then sub them for the race length but the problem that is getting me is the different constants for different times. Do I need the heavyside function or something to get this set-up.

 

[member 553083]

The solution to this question involves solving the equations of motion for both boats using the given information. The equation of motion for each boat is F = ma + bv, where F is the net force, m is the mass, a is the acceleration, b is the proportionality constant, and v is the velocity. Since the wind applies a constant force of 650N, we can substitute F = 650N in each equation. For Boat A, the equation of motion is 650N = 60kg(a) + 80Nsec/m(v) when v < 5m/s and 650N = 60kg(a) + 60Nsec/m(v) when v > 5m/s. For Boat B, the equation of motion is 650N = 60kg(a) + 100Nsec/m(v) when v < 6m/s and 650N = 60kg(a) + 50Nsec/m(v) when v > 6m/s.Once these equations are solved, they can be integrated with the initial conditions (initial velocity = 2m/s) to obtain the position of both boats over the 500m race. Whichever boat has a higher final position at the end of the race will be leading.