Question

Manuel is driving 60 miles per hour on a state road with a 65 mph speed limit. He sees a fallen tree up ahead and must come to a quick, complete stop.
a. What is his approximate reaction distance?
b. What is his approximate braking distance?
c. About how many feet does the car travel from the time he switches pedals until the car has completely stopped?

Answer 1

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.

How to analyzing reaction time in a braking event

The average reaction time ($t_{R}$) is 0.75 seconds. Manuel drives at constant velocity in the first 0.75 seconds, then he decelerates the vehicle.

a) The reaction distance ($x_{R}$), in meters, is found by the following expression:

$x_{R} = v_{o}\cdot t_{R}$ (1)

Where $v_{o}$ is the initial velocity, in feet per hour.

If we know that $v_{o} = 60\,\frac{mi}{h}$ ($v_{o} = 88\,\frac{ft}{s}$) and $t_{R} = 0.75\,s$, then the approximate reaction distance is:

$x_{R} = (88)\cdot (0.75)$

$x_{R} = 66\,ft$

The approximate reaction distance is 66 feet. $\blacksquare$

b) A normal braking has magnitudes of about 0.15 times the value of gravitational acceleration ($g = 32.174\,\frac{ft}{s^{2}}$). The approximate braking distance ($d$), in feet, is found by the following kinematic formula:

$d = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$ (2)

Where:

• $a$ - Deceleration rate, in feet per square second.
• $v$ - Final velocity, in feet per second.

If we know that $v_{o} = 88\,\frac{ft}{s}$ and $v_{o} = 0\,\frac{ft}{s}$, then the approximate braking distance is:

$d = \frac{\left(0\,\frac{ft}{s}\right)^{2}-\left(88\,\frac{ft}{s} \right)^{2}}{2\cdot \left(0.15\right)\cdot \left(-32.174\,\frac{ft}{s^{2}} \right)}$

$d = 802.304\,ft$

The approximate braking distance is 802.304 feet. $\blacksquare$

c) The total distance is the sum of distances found in a) and b). Then, the total distance is 868.304 feet. $\blacksquare$

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