Questions: In a performance test, each of two cars takes 9.7 s to accelerate from rest to 27 m / s. Car A has a mass of 1427 kg, and car B has a mass of 1898 kg. Find the net average force that acts on (a) car A and (b) car B during the test.

In a performance test, each of two cars takes 9.7 s to accelerate from rest to 27 m / s. Car A has a mass of 1427 kg, and car B has a mass of 1898 kg. Find the net average force that acts on (a) car A and (b) car B during the test.

Transcript text: In a performance test, each of two cars takes 9.7 s to accelerate from rest to $27 \mathrm{~m} / \mathrm{s}$. Car A has a mass of 1427 kg , and car B has a mass of 1898 kg . Find the net average force that acts on (a) car A and (b) car B during the test.

Solution

Solution Steps

Step 1: Identify the Given Data

We are given the following data for both cars:

Time to accelerate, $ t = 9.7 $ s
Final velocity, $ v = 27 $ m/s
Initial velocity, $ u = 0 $ m/s

For Car A:

Mass, $ m_A = 1427 $ kg

For Car B:

Mass, $ m_B = 1898 $ kg

Step 2: Calculate the Acceleration

Using the formula for acceleration: \[ a = \frac{v - u}{t} \] Substitute the given values: \[ a = \frac{27 \, \text{m/s} - 0 \, \text{m/s}}{9.7 \, \text{s}} = \frac{27}{9.7} \approx 2.7835 \, \text{m/s}^2 \]

Step 3: Calculate the Net Average Force for Car A

Using Newton's second law: \[ \overline{F}_A = m_A \cdot a \] Substitute the values for Car A: \[ \overline{F}_A = 1427 \, \text{kg} \cdot 2.7835 \, \text{m/s}^2 \approx 3971 \, \text{N} \]

Step 4: Calculate the Net Average Force for Car B

Using Newton's second law: \[ \overline{F}_B = m_B \cdot a \] Substitute the values for Car B: \[ \overline{F}_B = 1898 \, \text{kg} \cdot 2.7835 \, \text{m/s}^2 \approx 5282 \, \text{N} \]