12-13 Engineering Mechanics Dynamics 13th edition Tests reveal that a normal dri

Question

12-13 Engineering Mechanics Dynamics 13th edition Tests reveal that a normal driver takes about 0.75 s before he or she can react to a situation to avoid a collision. It takes about 3 s for a driver having 0.1% alcohol in his system to do the same. If such drivers are traveling on a straight road at 30 mph (44 ft/s) and their cars can decelerate at 2 ft/s^2, determine the shortest stopping distance d for each from the moment they see the pedestrians. Moral: If you must drink, please don't drive!

Explanation / Answer

stopping distance = distance travelled in 3s + distance required to stop the car = (3 * 44) + [ (44^2) / (2*a) ] where a is the deceleration provided by the brakes a = (v-u) / t = 44 / 0.75 = 58.6666 ft/s^2 Substituting the value of a, we get: safe stopping distance = (132) + [1936 / 58.6666] = 165 ft.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.