Question

A motorist traveling at 11 m/s encounters a deer in the road 45 m ahead.
If the maximum acceleration the vehicle’s brakes are capable of is −7 m/s
2, what is themaximum reaction time of the motorist that will allow her or him to avoid hitting the deer?
Answer in units of s.

Answer 1

Answer on Question #65341, Physics / Mechanics | Relativity |

A motorist traveling at $11\,\mathrm{m/s}$ encounters a deer in the road $45\,\mathrm{m}$ ahead. If the maximum acceleration the vehicle's brakes are capable of is $-7\,\mathrm{m/s^2}$, what is the maximum reaction time of the motorist that will allow her or him to avoid hitting the deer? Answer in units of s.

Solution

$$\begin{array}{l} v_0 = 11\,\mathrm{m/sec} \\ a = -7\,\mathrm{m/sec^2} \\ L = 45\,\mathrm{m} \\ t_{\text{react}} - ? \end{array}$$

First we find the braking distance $L_{\mathrm{br}}$. We use a formula for a velocity: $v = v_0 + a \cdot t$. In the moment of a complete stop $v = 0$, $v_0 + a \cdot t_{\mathrm{br}} = 0$. So a vehicle needs time to brake $t_{\mathrm{br}} = -v_0 / a = (-11) / (-7) = 1.6$ (sec). Therefore $L_{\mathrm{br}} = v_0 \cdot t_{\mathrm{br}} + (a \cdot t_{\mathrm{br}}^2) / 2 = 11 \cdot 1.6 - 7 \cdot 1.6^2 / 2 = 8.6$ (m).

Before the motorist must begin a braking, the vehicle can drive yet this distance: $L - L_{\mathrm{br}} = 45 - 8.6 = 36.4$ (m). But on this distance the car is moving at a constant velocity. So, now we find a time easily: $t_{\text{react}} = (L - L_{\mathrm{br}}) / v_0 = 36.4 / 11 = 3.3$ (sec).

Answer: 3.3 sec

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