Question

1. an active physics student group is studying an intersection. the width of the intersection is measured by pacing and is found to be approximately 15 - m wide. the yellow - light time for the intersection is 4 s. the speed limit on this road is 30 mi/h (approximately 15 m/s). the speed of an automobile decreases by 5 m/s every second during negative acceleration. assume that the people who are driving the automobiles have a reaction time of 1 s.

a) calculate the go zone using the math equation on the computer spreadsheet. use a calculator. to guide you, the first two steps are provided for you.
go zone=(velocity×yellow - light time)−width of intersection
gz = vt_y−w
gz=(15 m/s)(4 s)−15 m
gz=
b) calculate the stop zone using the math equation on the computer spreadsheet. use a calculator to help you.
stop zone=(velocity×reaction time)+velocity²/(2×negative acceleration)
sz = vt_r+ v²/2a
c) make a sketch of the intersection and label both the go zone and the stop zone. include the dimensions of the intersection and each zone.

Explanation:

Step1: Calculate GO Zone

We are given the formula $GZ = vt_y - w$, where $v = 15$ m/s, $t_y=4$ s and $w = 15$ m.
$GZ=(15\ m/s)\times(4\ s)-15\ m$
$GZ = 60\ m - 15\ m$
$GZ = 45\ m$

Step2: Calculate STOP Zone

We are given the formula $SZ=vt_r+\frac{v^{2}}{2a}$, where $v = 15$ m/s, $t_r = 1$ s and $a=- 5$ m/s².
First, calculate $vt_r$: $vt_r=(15\ m/s)\times(1\ s)=15\ m$.
Then, calculate $\frac{v^{2}}{2a}$: $\frac{v^{2}}{2a}=\frac{(15\ m/s)^{2}}{2\times(-5\ m/s^{2})}=\frac{225\ m^{2}/s^{2}}{- 10\ m/s^{2}}=-22.5\ m$.
$SZ=15\ m+22.5\ m$ (taking the magnitude as distance)
$SZ = 37.5\ m$

Answer:

a) $45$ m
b) $37.5$ m
c) (Sketch - not provided in text - but should be a drawing of an intersection with a width of 15 m. Mark a region starting from the start - of - intersection and extending 45 m as the GO Zone and a region starting from the end - of - intersection and extending 37.5 m as the STOP Zone)