Question

Consider a \(1400-\mathrm{kg}\) car cruising at constant speed of \(70 \mathrm{km} / \mathrm{s}\). Now the car starts to pass another car, by accelerating to \(110 \mathrm{km} / \mathrm{h}\) in \(5 \mathrm{s}\). Determine the additional power needed to achieve this acceleration. What would your answer be if the total mass of the car were only \(700 \mathrm{kg} ? \)

Short answer

Answer: The additional power needed for a 1400 kg car is 95,127.4 W. For a 700 kg car, the power required would be 47,563.7 W.

Step-by-step solution

Convert velocities to meters per second

Given initial velocity \(v_1 = 70 \mathrm{km/h}\), we convert it to meters per second: \(v_1 = 70 * \frac{1000}{3600} = 19.44 \mathrm{m/s}\) Given final velocity \(v_2 = 110 \mathrm{km/h}\), we convert it to meters per second: \(v_2 = 110 * \frac{1000}{3600} = 30.56 \mathrm{m/s}\)

Calculate acceleration

We are given the time taken to accelerate, \(t = 5 \mathrm{s}\). We can now calculate the acceleration, \(a\), using the formula: \(a = \frac{v_2 - v_1}{t} = \frac{30.56 - 19.44}{5} = 2.224 \mathrm{m/s^2}\)

Calculate force needed for acceleration

For mass \(m = 1400\,\mathrm{kg}\), we can use Newton's second law to calculate the force needed for this acceleration: \(F = m \cdot a = 1400 \cdot 2.224 = 3113.6 \,\mathrm{N}\)

Calculate the additional power needed

To find the power, \(P\), needed to achieve this acceleration, we use the formula: \(P = F \cdot v_2 = 3113.6 \cdot 30.56 = 95127.4 \,\mathrm{W}\) So, the additional power needed for the \(1400\,\text{kg}\) car is \(95,127.4\,\text{W}\)

Calculate power for the 700kg car

Repeat steps 3 and 4 using a mass of \(700\,\text{kg}\) to find the force and power needed for the acceleration: \(F_{700} = m \cdot a = 700 \cdot 2.224 = 1556.8\,\text{N}\) \(P_{700} = F_{700} \cdot v_2 = 1556.8 \cdot 30.56 = 47563.7\,\text{W}\) If the total mass of the car were only \(700\,\text{kg}\), the additional power needed would be \(47,563.7\,\text{W}\).