The power of an engine of car
mass of 1200 kg is 134.05 HP and its
efficiency is 90%. What is the
minimum time required to bring the car
from rest, up to a velocity of 30 ms-1?
To calculate the minimum time required to bring the car from rest to a velocity of 30 m/s, we need to consider the work done and the power of the engine.
First, let's calculate the work done. The work done is equal to the change in kinetic energy of the car. The formula for kinetic energy is:
KE = 0.5 * mass * velocity^2
Given:
Mass of car (m) = 1200 kg
Final velocity (v) = 30 m/s
Initial velocity is zero since the car is at rest.
Delta KE = KE final - KE initial = 0.5 * m * v^2 - 0.5 * m * 0^2
= 0.5 * 1200 kg * (30 m/s)^2 - 0.5 * 1200 kg * 0^2
= 0.5 * 1200 kg * (900 m^2/s^2)
= 540000 J
Now, let's calculate the power of the engine. The power is the rate at which work is done. It can be calculated using the formula:
Power = Work / Time
The power is given as 134.05 horsepower. To convert horsepower to watts, we use the conversion factor: 1 horsepower = 745.7 watts.
Power = 134.05 HP * 745.7 W/HP
Power = 100000 W (approximately)
Given that the efficiency of the engine is 90%, we can calculate the actual power output of the engine by multiplying it with the efficiency:
Actual Power Output = Power * Efficiency
Actual Power Output = 100000 W * 0.9
Actual Power Output = 90000 W
Now, we can calculate the time required using the equation for power:
Time = Work / Power
Time = 540000 J / 90000 W
Time = 6 seconds
Therefore, the minimum time required to bring the car from rest to a velocity of 30 m/s is 6 seconds.