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Chapter 2: Problem 63

In the problems, please assume the free-fall acceleration $g=9.80 \mathrm{m} / \mathrm{s}^{2}$ unless a more precise value is given in the problem statement. Ignore air resistance. A car traveling at \(29 \mathrm{m} / \mathrm{s}(65 \mathrm{mi} / \mathrm{h})\) runs into a bridge abutment after the driver falls asleep at the wheel. (a) If the driver is wearing a seat belt and comes to rest within a 1.0 -m distance, what is his acceleration (assumed constant)? (b) A passenger who isn't wearing a seat belt is thrown into the windshield and comes to a stop in a distance of \(10.0 \mathrm{cm} .\) What is the acceleration of the passenger?

Short Answer

Expert verified
Based on the calculations, the driver's acceleration is -420.5 m/s², while the passenger's acceleration is -4205 m/s² as they come to a stop after the car crashes into a bridge.

Step by step solution

01

Calculate the driver's acceleration

First, we will plug in the values for the driver, with the stopping distance \(s=1.0\) m. Using the equation of motion, \(0^2 = 29^2 + 2a(1.0)\). Solving for the acceleration \(a\): \(a = \frac{0 - 29^2}{2(1.0)}\) \(a = \frac{-841}{2}\) \(a = -420.5 \, \mathrm{m/s^2}\) Thus the driver's acceleration is -420.5 m/s² (negative indicates deceleration).
02

Calculate the passenger's acceleration

Now, we will plug in the values for the passenger with the stopping distance \(s=0.1\) m. Using the equation of motion again, \(0^2 = 29^2 + 2a(0.1)\). Solving for the acceleration \(a\): \(a = \frac{0 - 29^2}{2(0.1)}\) \(a = \frac{-841}{0.2}\) \(a = -4205 \, \mathrm{m/s^2}\) The passenger's acceleration is -4205 m/s² (negative indicates deceleration).

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