Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 845-kg electric car be able to supply to do the following?

Question

(a) accelerate from rest to 25.0 m/s in 1.00 min

(b) climb a 200-m high hill in 2.00 min at a constant 25.0 m/s speed while exerting 440 N of force to overcome air resistance and friction

Answer 1

a. Work = Change in kinetic energy:
Work = 0.5M*V^2-0.5M*Vo^2, Vo = 0, Work = 0.5M*V^2 = 0.5*845*25^2 = 264,063 J.
Po = W/t = 264063/60 = 4401 J/s. = 4401 Watts = Power out.

Pi = 4401/0.95 = 4633 Watts = Power in.

Pi = 12*I = 4633, I = 4633/12 = 386 Amps.