Physics: Along a straight road through town, there are three speed-limit signs.
ID: 2227751 • Letter: P
Question
Physics: Along a straight road through town, there are three speed-limit signs. They occur in the following order: 66 , 37 , and 17 mi/h, with the 37 -mi/h sign located midway between the other two. Obeying these speed limits, the smallest possible time tA that a driver can spend on this part of the road is to travel between the first and second signs at 66 mi/h and between the second and third signs at 37 mi/h. More realistically, a driver could slow down from 66 to 37 mi/h with a constant deceleration and then do the same thing from 37 to 17 mi/h. This alternative requires a time tB. Find the ratio tB/tAExplanation / Answer
The route is divided into two equal parts of unknown distance x. Following plan A, speed
is constant between signs. Therefore, we may use the equation x = vt. Solving for t gives
us t = x/v. So, between the rst two signs, the time is x/66 and between the next two
signs, the time is x/37.
The total time under plan A is tA = x/66 + x/37.
For plan B, we must use the equation x =1/2 (v0+v)t. Solving this equation for t gives us t =2x/(v0 +v). The total amount of time spent in plan B is tB = 2x/(66+ 37) + 2x/(37+ 17).The ratio of times is
tB/tA = (2x/103 + 2x/54)/(x/66 + x/37)
=(2/103 +2/54)/(1/66 +1/37)
=1.338