MEI (OCR) M4 June 2006 Q3

The question may be seen here: https://pmt.physicsandmathstutor.com/download/Maths/A-level/M4/Papers-OCR-MEI/Combined%20QP%20-%20M4%20OCR%20MEI.pdf
i) Let A = 4E-4 W, B=1E4 m/s^2 s.t. P = mAV*(B+V^2).Recognise power = rate of work done by net force P = FV=mdV/dtVHence dV/dt = A(B+V^2). Use chain rule s.t. dV/dt = dV/dxV and separate variables, integrating (dV/dt)V/(B+V^2) = A wrt x, giving V(x) = sqrt(B[exp(2Ax)-1]) noting that V(0)=0. Solve for x, V=80 gives x=618.4<900m available. Also dV/dx>0 for all V!=0 so V is monotonically increasing and this is the only solution.
ii) From (i) we have (dx/dt)/sqrt(exp(2Ax)-1)=sqrt(B). Integrating wrt t, making the substitution u=sqrt(exp(2Ax)-1) gives V(t) = sqrt(B)tan(Asqrt(B)t), again using V(0)=0. This is unbounded towards t=pi/(2Asqrt(B)) and is therefore unphysical for t>39s.
iii) Solving the above model for V=80m/s reveals that the model is out of bounds and the given revision with constant power is required. Plugging in t=11s (model limit) to the earlier equations gives a limit V=47.1m/s, P/m=230.2W/kg, x=250.4m. At constant power, the equation of motion is P/m = VdV/dt = constant. Integrating wrt t gives V(t) = sqrt(2*(Pt/m+K)), where K =47.1^2/2-230.211 from the boundary condition, i.e. V(t)=sqrt(460.4t-2846). Solving for 80m/s gives t=20.1s (i.e. a solution exists). The corresponding distance travelled may be found by solving the DE V=dx/dt, i.e. integrating the above wrt time, giving x-x(11)=2sqrt(2)/(3P/m)(Pt/m+K)^1.5 on t=[11,20.1]. Thence, x(V=80) is 841.9m<900m available and take off is successful.
Working and diagrams as attachment.