Using Newton's second law, the net force on the car is given by:

\(F - R = ma\)

At point A, the force equation is:

\(F_A - 800 = 600a_A\)

The power of the engine is 40 kW, so:

\(F_A = \frac{40000}{25} = 1600 \text{ N}\)

Substituting into the force equation at A:

\(1600 - 800 = 600a_A\)

\(800 = 600a_A\)

\(a_A = \frac{800}{600} = \frac{4}{3} \text{ m s}^{-2}\)

At point B, the acceleration is half of that at A:

\(a_B = \frac{1}{2}a_A = \frac{2}{3} \text{ m s}^{-2}\)

Using the force equation at B:

\(\frac{40000}{v_B} - 800 = 600 \left(\frac{2}{3}\right)\)

\(\frac{40000}{v_B} - 800 = 400\)

\(\frac{40000}{v_B} = 1200\)

\(v_B = \frac{40000}{1200} = 33.3 \text{ m s}^{-1}\)