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Chapter 5: Problem 75

What's the fastest way to make a U-turn at constant speed? Suppose that you need to make a \(180^{\circ}\) turn on a circular path. The minimum radius (due to the car's steering system) is \(5.0 \mathrm{m},\) while the maximum (due to the width of the road) is \(20.0 \mathrm{m}\). Your acceleration must never exceed \(3.0 \mathrm{m} / \mathrm{s}^{2}\) or else you will skid. Should you use the smallest possible radius, so the distance is small, or the largest, so you can go faster without skidding, or something in between? What is the minimum possible time for this U-turn?

Short Answer

Expert verified
Answer: The optimal radius is 20.0m and the minimum possible time for the U-turn is \(\dfrac{20\pi}{\sqrt{60}}\) seconds.

Step by step solution

01

Calculate the centripetal acceleration limitation

Using the formula for centripetal acceleration: \(a_c = \dfrac{v^2}{r}\). We have a limit for the centripetal acceleration, \(a_c = 3.0 \mathrm{m/s^2}\). We'll solve this equation for \(v\) using the given values of \(r\) and find the minimum and maximum allowable speeds for each radius. For r=5.0m: \(3.0 = \dfrac{v^2}{5.0}\), \(v = \sqrt{15.0}\) For r=20.0m: \(3.0 = \dfrac{v^2}{20.0}\), \(v = \sqrt{60.0}\)
02

Calculate the distance traveled for each radius

The distance traveled for each radius can be calculated using the formula for the arc length of a circle. We are given that the turn angle is \(180^{\circ}\), which is \(\pi\) radians. The formula for the arc length is: \(s = r \times \theta\) For r=5.0m: \(s_{min} = 5.0 \times \pi = 5\pi \mathrm{m}\) For r=20.0m: \(s_{max} = 20.0 \times \pi = 20\pi \mathrm{m}\)
03

Calculate the time required for U-turn for each radius

Now that we have the distances and velocities for each radius, we can calculate the time using the formula: \(t = \dfrac{s}{v}\) For r=5.0m: \(t_{min} = \dfrac{5\pi}{\sqrt{15}}\) For r=20.0m: \(t_{max} = \dfrac{20\pi}{\sqrt{60}}\)
04

Compare the times and choose the optimal radius

Comparing the times for both the minimum and maximum radius, we can see that the minimum radius yields a longer time: \(t_{min} = \dfrac{5\pi}{\sqrt{15}} > \dfrac{20\pi}{\sqrt{60}} = t_{max}\) Thus, using the largest possible radius (20.0m) will result in the fastest U-turn without skidding.
05

Calculate the minimum possible time for the U-turn

Now that we have determined that the maximum radius provides the fastest U-turn, we can use the calculated time value for this maximum radius case: \(t_{fastest} = \dfrac{20\pi}{\sqrt{60}}\) The minimum possible time for this U-turn is \(\dfrac{20\pi}{\sqrt{60}}\) seconds.

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