To stop a car, first you require a certain reaction time to begin braking; then

Question

To stop a car, first you require a certain reaction time to begin braking; then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 53.8 m when its initial speed is 77.9 km/h, and 23.1 m when its initial speed is 47.1 km/h. What are (a) your reaction time and (b) the magnitude of the deceleration?

Explanation / Answer

Let Tr = reaction time; t = deceleration time, case #1; T = deceleration time, case #2 Case #1: V1 = 75.8km/hr·Tr·(1000m/km)·(hr/3600s) = 21 m/s Case #2: V2 = 50.4km/hr·Tr·(1000m/km)·(hr/3600s) = 14 m/s I'm going to drop units, since we're consistent now (meters and seconds). For both cases, total distance = distance covered during Tr + distance covered during braking. S = Vo·Tr + (Vo·Tb + ½a(Tb)²) where Tb is the braking time, t and T respectively 1: 52.3 = 21·Tr + 21·t + ½at² 2: 24.9 = 14·Tr + 14·T + ½aT² Solve both for Tr (reaction time): 1: Tr = (52.3 - 21t - ½at²) / 21 -----> #1 2: Tr = (24.9 - 14T - ½aT²) / 14 -------> #2 These equal, so (52.3 - 21t - ½at²) / 21 = (24.9 - 14T - ½aT²) / 14 mult by 21 (52.3m -21t - ½at²) = (24.9 -14T - ½aT²) · 1.5 ------> #3 Two eqns, 3 unknowns (a, t, T). Need more eqns. How about Toricelli's? v² = 0 = u² - 2as No good, we add the unknown "s" How about velocity / time? V = Vo +at 1: V = 0 = Vo + at = 21m/s + at ? a = -21m/s / t ------> #4 2: V = 0 = 14m/s + aT ? a = -14m/s / T --------> #5 Since a = a, -21m/s · T = -14m/s · t T = (2/3)t -----> #6 Use #4 (for a) and #6 (for T) and plug into #3: (52.3 -21t - ½at²) = (24.9 -14T - ½aT²) · 1.5 ------> #3 (52.3 - 21t - ½(-21 / t)t²) = (24.9 -14(2/3)t - ½(-21 / t)((2/3)t)²) · 1.5 52.3 - 21t - 10.5·t = (24.9 - 9.3t - 10.5·t·(4/9) ) · 1.5 52.3 - 31.5·t = 37.3 - 21·t 15 = 10.5 · t t = 1.43 s T = (2/3)t = 0.95 s a = -21m/s / t = -21m/s / 1.43s = 14.7 m/s² ? (b) 1: Tr = (52.3 - 21t - ½at²) / 21 = (52.3 - 21(1.43) - ½(14.7)1.43²) / 21 = 0.34 s ? (a) 2: Tr = (24.9 - 14T - ½aT²) / 14 = (24.9 - 14(0.95) - ½(14.7)0.95²) / 14 = 0.35 s v Well, I got done with this and thought "that was ridiculous". There must be a better way. I found this answer: http://answers.yahoo.com/question/index;… This answerer uses Torricelli's eqn in an economical way by acknowledging that the distance over which the acceleration occurs = the total distance - distance covered during reaction time, or v² - u² = v² = 2as So for case 1: (21m/s)² = 2a(52.3m - 21m/s·Tr) and for case 2: (14m/s)² = 2a(24.9m - 14m/s·Tr) Solve both for 2a and set equal to each other and you get (21m/s)² / (52.3m - 21m/s·Tr) = (14m/s)² / (24.9m - 14m/s·Tr) You'll find that Tr = 0.34s works for this.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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