C. 8 to 10 times.
Let suppose that vehicle decelerates uniformly. and in both cases, vehicles decelerates at same rate. In this case, we are asked to determine the Distance ratio between two Kinematic scenarios We can estimate the proportion based on relationships inherent to Kinematic formulas associated with uniform Accelerated Motion, of which we proceed to use the following expression:
[tex]v^{2} = v_{o}^{2} + 2\cdot a\cdot x[/tex]
Where:
[tex]v_{o}[/tex] - Initial velocity, in miles per hour.
[tex]a[/tex] - Acceleration, in miles per square hour.
[tex]v[/tex] - Final velocity, in miles per hour.
[tex]x[/tex] - Travelled distance, in miles.
As we can notice difference between the squares of final and initial Velocities is directly proportional to Time. Hence, we can obtain the following time ratio:
[tex]r_{s} = \frac{\left(0\,\frac{mi}{h}\right)^{2} - \left(60\,\frac{mi}{h}\right)^{2} }{\left(0\,\frac{mi}{h}\right)^{2}-\left(20\,\frac{mi}{h}\right)^{2} }[/tex]
[tex]r_{s} = 9[/tex]
Compare to driving at 20 miles per hour, driver will take 9 times more the distance to stop at 60 miles per hour. (Answer: C)
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