Question

A car can be brakes to stop from 60 mi/h in 43 m.
a) what is the magnitude of the acceleration in SI units and “g units”? Assume that acc is constant.
b) what is the stopping time? If your reaction time T for braking is 400 ms to how many “reaction times” does the stopping time correspond?

Answer 1

Answer on Question#82112 - Physics - Other

A car can be brakes to stop from 60 mi/h in 43 m.

a) what is the magnitude of the acceleration in SI units and "g units"? Assume that acc is constant.

b) what is the stopping time? If your reaction time T for braking is 400 ms to how many "reaction times" does the stopping time correspond?

Solution:

Since 1 mile = 1,60934 km, we obtain

$$v_i = 60 \frac{\mathrm{mi}}{\mathrm{h}} = 60 \cdot 1,60934 \frac{\mathrm{km}}{\mathrm{h}} = 96.56 \frac{\mathrm{km}}{\mathrm{h}} = 26.82 \frac{\mathrm{m}}{\mathrm{s}}$$

The initial $v_i$ and final $v_f$ speeds are related to the stopping length $l$ and acceleration $a$ in the following way

$$v_f^2 - v_i^2 = 2al$$

Thus

$$a = \frac{v_f^2 - v_i^2}{2l}$$

Since $l = 43$ m and $v_f = 0\frac{\mathrm{m}}{\mathrm{s}}$, we obtain

$$a = \frac{\left(0 \frac{\mathrm{m}}{\mathrm{s}}\right)^2 - \left(26.82 \frac{\mathrm{m}}{\mathrm{s}}\right)^2}{2 \cdot 43 \mathrm{m}} = -8.36 \frac{\mathrm{m}}{\mathrm{s}^2}$$

Since $g = 9.80665 \, \mathrm{m/s}^2$, the acceleration in g units is given by

$$a = \frac{-8.36 \frac{\mathrm{m}}{\mathrm{s}^2}}{9.80665 \frac{\mathrm{m}}{\mathrm{s}^2}} = 0.85g$$

The stopping time is given by

$$t = \sqrt{\frac{2l}{|a|}} = \sqrt{\frac{2 \cdot 43 \mathrm{m}}{8.36 \frac{\mathrm{m}}{\mathrm{s}^2}}} = 3.2 \mathrm{s}$$

Thus the ratio of the stopping time $t$ to reaction time ($T = 400 \, \mathrm{ms} = 0.4 \, \mathrm{s}$) is

$$\frac{t}{T} = \frac{3.2 \mathrm{s}}{0.4 \mathrm{s}} = 8$$

Answer:

$a = -8.36 \frac{\mathrm{m}}{\mathrm{s}^2} = 0.85g, t = 3.2 \, \mathrm{s} = 8T.$

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