Answer to Question #91635 in Mechanics | Relativity for Marisol Cuevas

Question

The maximum straight-line deceleration of a racing car under braking is 5m/s^-2. What is the minimum stopping distance of the car from a velocity of 100km/h? What does the distance become if the velocity has twice this value?

Answer 1

Assuming that deceleration rate is constant, the stopping distance can be calculated in the way as follows:

$$s = \frac{v^2}{2 a}$$

The minimal stopping distance occurs in the case when the deceleration is maximal. Hence,

$$s_{min} = \frac{v^2}{2 a_{max}}$$

Taking into account that 100 km/h = 27.8 m/s and substituting the numerical values, we obtain:

$$s_{min} = \frac{27.8^2}{2 \cdot 5} \approx 77.3 \, m$$

If the initial velocity doubles, i.e.

$$v_2 = 2v$$

then

$$s_2 = \frac{v_2^2}{2 a_{max}} = \frac{(2v)^2}{2 a_{max}}=4 \frac{v^2}{2 a_{max}} =4 s_{min} \approx 309.1 \, m$$

Answer: 77.3 m; 309.1 m.