Question

Maria is racing her bike down the cobblestone sidewalk. She is riding at a constant velocity of 10.2m/s. Suddenly a group of people step out in front of her. They are 6 meters in front and she needs to stop. The coefficient of friction between the ground and Maria's tire is 0.35. The mass of her + her bike is 130kg. She slams the brakes with 800N of force. Will she be able to stop in time? Or will she run over the people?

Answer 1

Question#18717

Maria is racing her bike down the cobblestone sidewalk. She is riding at a constant velocity of 10.2m/s. Suddenly a group of people step out in front of her. They are 6 meters in front and she needs to stop. The coefficient of friction between the ground and Maria's tire is 0.35. The mass of her + her bike is 130kg. She slams the brakes with 800N of force. Will she be able to stop in time? Or will she run over the people?

Solution:

Let:

$$v = 10.2 \text{ m/s}$$ $$S = 6 \text{ m}$$ $$m = 150 \text{ kg}$$ $$F = 800 \text{ N}$$ $$k = 0.35$$

The brake distance is:

$$S_{\text{brake}} = vt - \frac{1}{2}at^2, \quad v = at, t = \frac{v}{a}$$ $$S_{\text{brake}} = \frac{v^2}{a} - \frac{v^2}{2a} = \frac{v^2}{2a}, \quad \text{were a is the brake acceleration}$$ $$a = \frac{F}{m}$$ $$S_{\text{brake}} = \frac{mv^2}{2F} = \frac{150 \times 10.2^2}{2 \times 800} = 9.75 \text{ m}$$

Such as the brake distance (without friction factor) is more than distance needs to stop, she will run over the people. With friction factor the brake way will be more.