Question

in a vertical dive, a peregrine falcon can accelerate at 0.6 times the free-fall acceleration g (that is, at 0.6g) in reaching a speed of about 111 m/s. if a falcon pulls out of a dive into a circular arc at this speed and can sustain a radial acceleration of 0.6g, what is the minimum radius r of the turn? r = \boxed{\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_} km

Explanation:

Step1: Recall radial acceleration formula

The formula for radial (centripetal) acceleration is \( a_{r}=\frac{v^{2}}{R} \), where \( a_{r} \) is radial acceleration, \( v \) is speed, and \( R \) is the radius of the circular path. We know \( a_{r} = 0.6g \), \( v = 111\space m/s \), and \( g = 9.8\space m/s^{2} \). We need to solve for \( R \). Rearranging the formula gives \( R=\frac{v^{2}}{a_{r}} \).

Step2: Substitute \( a_{r} = 0.6g \) into the formula

Substitute \( a_{r}=0.6g \) into \( R=\frac{v^{2}}{a_{r}} \), we get \( R = \frac{v^{2}}{0.6g} \).

Step3: Plug in the values of \( v \), \( g \)

Now plug \( v = 111\space m/s \), \( g = 9.8\space m/s^{2} \) into the formula:
\[ \begin{align*} R&=\frac{(111)^{2}}{0.6\times9.8}\\ &=\frac{12321}{5.88}\\ &\approx2095.41\space m \end{align*} \]

Step4: Convert meters to kilometers

Since \( 1\space km = 1000\space m \), divide the result by 1000: \( R=\frac{2095.41}{1000}\approx2.10\space km \) (rounded to two decimal places) or more precisely, let's recalculate the division:

First, \( \frac{12321}{5.88}= \frac{1232100}{588}=\frac{1232100\div12}{588\div12}=\frac{102675}{49}\approx2095.408\space m \). Then convert to km: \( 2095.408\space m=2095.408\times10^{- 3}\space km = 2.095408\space km\approx2.10\space km \) (or if we keep more precision, let's check the calculation again:

\( v = 111\space m/s \), so \( v^{2}=111\times111 = 12321\space m^{2}/s^{2} \)

\( 0.6g=0.6\times9.8 = 5.88\space m/s^{2} \)

\( R=\frac{12321}{5.88}=\frac{1232100}{588}=2095.408163\space m \)

Convert to km: \( 2095.408163\space m=\frac{2095.408163}{1000}\space km = 2.095408163\space km\approx2.10\space km \) (or approximately \( 2.1\space km \) if we round to one decimal place). Wait, maybe we should do the calculation more accurately:

\( 12321\div5.88 \):

\( 5.88\times2000 = 11760 \)

\( 12321 - 11760=561 \)

\( 561\div5.88\approx95.408 \)

So total \( R\approx2000 + 95.408=2095.408\space m = 2.095408\space km\approx2.10\space km \) (or \( 2.1\space km \) depending on significant figures. The given values: \( 0.6g \) (one significant figure? Wait, 0.6g could be considered as one or two? Wait, 0.6 is one significant figure? No, 0.6 is one significant figure? Wait, 0.6 has one significant figure, 111 has three, 9.8 has two. So when multiplying/dividing, the result should have the least number of significant figures, which is one? But that seems too strict. Maybe 0.6g is considered as two significant figures (0.60g? No, the problem says 0.6g). Alternatively, maybe the problem expects us to use \( g = 9.8\space m/s^{2} \) (two significant figures), 0.6 (one), 111 (three). Hmm, maybe the problem expects us to proceed with the numbers as given. Let's recalculate:

\( R=\frac{v^{2}}{0.6g}=\frac{111^{2}}{0.6\times9.8}=\frac{12321}{5.88}\approx2095.4\space m = 2.0954\space km\approx2.10\space km \) (or \( 2.1\space km \)). Wait, maybe I made a mistake in the formula. Wait, radial acceleration is centripetal acceleration, so \( a_{c}=\frac{v^{2}}{R} \), so \( R = \frac{v^{2}}{a_{c}} \). Here, \( a_{c}=0.6g \), so that's correct.

Wait, let's check the calculation again:

\( 111^2 = 12321 \)

\( 0.6\times9.8 = 5.88 \)

\( 12321\div5.88 = 12321\div\frac{588}{100}=12321\times\frac{100}{588}=\frac{1232100}{588} \)

Divide numerator and denominator by 12: \( \frac{102675}{49}\approx2095.408 \) meters. Convert to kilometers: \( 2095.408\div1000 = 2.095408 \) kilometers, which is approximately \( 2.10\space km \) (or \( 2.1\space km \) if we round to two significant figures, since 0.6 has one, 9.8 has two, 111 has three. The least is one, but that would be 2 km, which seems too rough. Maybe the problem expects us to use \( g = 9.81\space m/s^{2} \) for more accuracy. Let's try that:

\( 0.6\times9.81 = 5.886 \)

\( R=\frac{12321}{5.886}\approx\frac{12321}{5.886}\approx2093.27\space m = 2.09327\space km\approx2.09\space km \) or \( 2.1\space km \).

But the standard answer is likely around \( 2.1\space km \) or \( 2.10\space km \). Let's see, maybe the calculation is:

\( R=\frac{v^{2}}{0.6g}=\frac{111^{2}}{0.6\times9.8}=\frac{12321}{5.88}\approx2095\space m = 2.095\space km\approx2.10\space km \) (rounded to two decimal places) or \( 2.1\space km \) (rounded to one decimal place).

Answer:

\( \approx2.10 \) (or \( 2.1 \)) km (depending on rounding, the precise value is approximately \( 2.10\space km \))