A commuter travels to work along a road with many traffic signals. After stoppin

Question

A commuter travels to work along a road with many traffic signals. After stopping at one red light the commuter accelerates at 2 m/s2. Just as he passes a speed camera, he notices that the next traffic light has turned red and he must stop, which he does by reversing his acceleration and now decelerating at a constant rate of 2 m/s2; coming to a stop just at the traffic light. The distance between the two traffic lights is 800 m.

A. How does the time he is speeding up compare to the time he is slowing down? Same, You cannot determine from the information given, More time speeding up, or More time slowing down. Explain your reasoning. B. How much time does he take to get from the first traffic light to the second? (in seconds) C.What was his maximum speed? (in m/s) D. The speed limit on the road is 45 mi/h. If he is going more than about 15 mi/h faster than that, hell get a traffic ticket in the mail. Do you expect him to get one? No, It cannot be determined from the information, or yes. Explain your reasoning.

Explanation / Answer

while speeding up

commuter travels a distance with acceleration a = + 2m/s^2 for t1 = t

x = 0.5*2*t^2

x = t^2 >..........(1)

speed after t time , v = 2*t

while decelerating

travels a distance 800-x in t2 time and comes to rest

V2 = 0

v2^2-v^2 = 2*a*(800-x)

4t^2 = 2*2*(800-x)

t^2 = 800 -x ........(2)

from 1 & 2

2t^2 = 800

t^2 = 400

t = 20 s

x = 400 m

while decelerating

v2 = v + a*t2

0 = 2*t - 2*t2

t2 = t = 20 s

part(A)

same

+++++++++++++

part(B)

t = 20 s

part(C)

v = 2*t = 2*20 = 40 m/s

++++++++++++

part(D)

speed limit to get a ticket = 45 + 15 = 60 mi/h = 26.8 m/s

v > 26.8 m/s

he will get a mail ticket

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