Final answer:
The power required to accelerate a 1500 kg car from 12.6 m/s to 27.8 m/s in 4.7 seconds is approximately 98.2 kilowatts or 131.55 horsepower.
Step-by-step explanation:
To determine the power required for a 1500 kg car to accelerate from 12.6 m/s to 27.8 m/s in 4.7 seconds, we can use the work-energy principle and the definition of power. Power is the rate at which work is done, or energy is transferred. Work done on the car is equal to the change in its kinetic energy.
First, we calculate the work done (W):
W = ½ m (v_f^2 - v_i^2)
where m is the mass of the car, v_f is the final velocity, and v_i is the initial velocity.
Plugging in the values:
W = ½ × 1500 kg (× (27.8 m/s)^2 - (12.6 m/s)^2) = ½ × 1500 kg × (773.84 - 158.76) = ½ × 1500 kg × 615.08 = 461310 J (joules)
Now we calculate the power (P):
P = W / t
where t is the time over which the work is done.
So, P = 461310 J / 4.7 s = 98172.34 W
Thus, the power required is about 98.2 kilowatts (kW). To convert this to horsepower (hp), we use the conversion factor (1 hp = 746 W).
Power in hp = 98172.34 W / 746 W/hp ≈ 131.55 hp
The car requires approximately 98.2 kW or 131.55 hp to achieve the desired acceleration.