the ODE is simple and it can be split into 2 parts:
1000*dV/dt+cd*1.225*V^2*A+200 - T
a basic 1st degree ODE that doesn't really need any solver to solve it, its just the integration of a polynomial:
The Torque applied
How does the Torque and its integral look like?
clear all;close all;clc
cd = 0.5;
T=10000*(t.*(t<1) + (t>=1).*(t<=50));
axis([-10 130 -100 10500])
title(' Torque and int(Torque)')
axis([-5 125 -100 5e6])
An example of real Torque is available in page 26 of Balaji Kamalammannan's Master Degree thesis
'Modelling and Simulation of Vehicle Kinematics and Dynamics (WITH MATLAB)'
Just split the needed integration
int(f1 + f2)= inf(f1) + int(f2)
syms dvdt v
(49*v^3)/240000 + v/5
T1 being cumsum(T)
Since the integral of the Torque is already null at v=0, let's find the integration constant to meet the initial condition
v=(49*v^2)/80000 + 1/5+T1 + kv0
fv3=@(u) (49*u^2)/80000 + 1/5+T1k+ kv1 - u
fv3=@(u) (49*u^2)/80000 + 1/5+T1k+ kv1 - u;
axis([0 100 -100 900])
As expected the dominant term of the velocity, with the equation you use, is the Torque.
Torque is roughly proportional to fuel (or battery charge) consumption.
But your equation is for an ideal model, with the vehicle on flat terrain. In reality, only with certain minimum downhill angle, and not even with brand new bearings and everything working well, air drag, friction continuously reduces gained velocity thus further torque is required to keep speed constant.
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