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a) The approximate reaction distance is 66 feet.
b) The approximate braking distance is 802.304 feet.
c) The total distance is 868.304 feet.
The average reaction time ([tex]t_{R}[/tex]) is 0.75 seconds. Manuel drives at constant velocity in the first 0.75 seconds, then he decelerates the vehicle.
a) The reaction distance ([tex]x_{R}[/tex]), in meters, is found by the following expression:
[tex]x_{R} = v_{o}\cdot t_{R}[/tex] (1)
Where [tex]v_{o}[/tex] is the initial velocity, in feet per hour.
If we know that [tex]v_{o} = 60\,\frac{mi}{h}[/tex] ([tex]v_{o} = 88\,\frac{ft}{s}[/tex]) and [tex]t_{R} = 0.75\,s[/tex], then the approximate reaction distance is:
[tex]x_{R} = (88)\cdot (0.75)[/tex]
[tex]x_{R} = 66\,ft[/tex]
The approximate reaction distance is 66 feet. [tex]\blacksquare[/tex]
b) A normal braking has magnitudes of about 0.15 times the value of gravitational acceleration ([tex]g = 32.174\,\frac{ft}{s^{2}}[/tex]). The approximate braking distance ([tex]d[/tex]), in feet, is found by the following kinematic formula:
[tex]d = \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex] (2)
Where:
If we know that [tex]v_{o} = 88\,\frac{ft}{s}[/tex] and [tex]v_{o} = 0\,\frac{ft}{s}[/tex], then the approximate braking distance is:
[tex]d = \frac{\left(0\,\frac{ft}{s}\right)^{2}-\left(88\,\frac{ft}{s} \right)^{2}}{2\cdot \left(0.15\right)\cdot \left(-32.174\,\frac{ft}{s^{2}} \right)}[/tex]
[tex]d = 802.304\,ft[/tex]
The approximate braking distance is 802.304 feet. [tex]\blacksquare[/tex]
c) The total distance is the sum of distances found in a) and b). Then, the total distance is 868.304 feet. [tex]\blacksquare[/tex]
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