Question

The engine of a \(1500-\mathrm{kg}\) automobile has a power rating of \(75 \mathrm{kW}\). Determine the time required to accelerate this car from rest to a speed of \(100 \mathrm{km} / \mathrm{h}\) at full power on a level road. Is your answer realistic?

Short answer

Answer: 15.4 seconds

Step-by-step solution

Convert the speed to m/s

To convert the speed from km/h to m/s, we can use the following conversion: \(1 \mathrm{km/h} = \frac{1000}{3600} \mathrm{m/s}\) So, \(v = 100 \mathrm{km/h} \times \frac{1000}{3600} \mathrm{m/s} = 27.8 \mathrm{m/s}\)

Find the force exerted by the engine

Given the power rating, \(P = 75 \mathrm{kW}\), and using the equation \(P = F \times v\), we can find the force exerted by the engine: \(F = \frac{P}{v}\) Substitute values of \(P\) and \(v\): \(F = \frac{75 \times 10^3 \mathrm{W}}{27.8 \mathrm{m/s}} = 2698 \mathrm{N}\)

Calculate the acceleration of the car

Using the equation for force, \(F = ma\), we can find the car's acceleration: \(a = \frac{F}{m}\) Substitute values of \(F\) and \(m\): \(a = \frac{2698 \mathrm{N}}{1500 \mathrm{kg}} = 1.8 \mathrm{m/s^2}\)

Calculate the time required to reach the final speed

Using the equation \(v = a \times t\), we can find the time required: \(t = \frac{v}{a}\) Substitute values of \(v\) and \(a\): \(t = \frac{27.8 \mathrm{m/s}}{1.8 \mathrm{m/s^2}} = 15.4 \mathrm{s}\)

Answer and Realism Check

The time required to accelerate the car from rest to a speed of \(100 \mathrm{km/h}\) at full power on a level road is \(15.4 \mathrm{seconds}\). This answer seems realistic for a car with a \(75\mathrm{kW}\) engine.