Question

In a performance test, each of two cars takes 8.3 s to accelerate from rest to 29 m/s. Car A has a mass of 1361 kg, and car B has a mass of 1871 kg. Find the net average force that acts on (a) car A and (b) car B during the test.

Answer 1

In a performance test, each of two cars takes 8.3 s to accelerate from rest to 29 m/s. Car A has a mass of 1361 kg, and car B has a mass of 1871 kg. Find the net average force that acts on (a) car A and (b) car B during the test.

**Solution:**

$$\begin{array}{l} V = 29 \frac{\mathrm{m}}{\mathrm{s}} - \text{velocity of the cars}; \\ t = 8.3 \mathrm{s} - \text{time required to reach the speed } V \\ \mathrm{m}_{\mathrm{A}} = 1361 \mathrm{kg} - \text{mass of the car A}; \\ \mathrm{m}_{\mathrm{B}} = 1871 \mathrm{kg} - \text{mass of the car B}; \\ \mathrm{F}_{\mathrm{A}}, \mathrm{F}_{\mathrm{B}} - \text{net forces that acts on cars A and B during the test}. \end{array}$$

First, we can find the acceleration of the car. Rate equation along the X axis:

$$\begin{array}{l} V = a t \\ a = \frac{V}{t} \end{array}$$

Newton's second law for the car:

$$F = m a = \frac{m V}{t} \Rightarrow$$ $$F_{A} = \frac{m_{A} V}{t} = \frac{1361 \mathrm{kg} \cdot 29 \frac{\mathrm{m}}{\mathrm{s}}}{8.3 \mathrm{s}} = 4755 \mathrm{N}$$ $$F_{B} = \frac{m_{B} V}{t} = \frac{1871 \mathrm{kg} \cdot 29 \frac{\mathrm{m}}{\mathrm{s}}}{8.3 \mathrm{s}} = 6537 \mathrm{N}$$

Answer: a) 4755N

b) 6537N