A person driving her car at 50 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow light lasts only 2.0 s before turning to red, and she is 30 m away from the near side of the intersection. The intersection is 15 m wide. Her car's maximum deceleration is -6.6 m/s2, whereas it can accelerate from 50 km/h to 70 km/h in 6.0 s. Ignore the length of her car and her reaction time.If she hits the brakes, how far will she travel before stopping?
+37084 This question has a LOT of extraneous information which does not apply to the question asked.....perhaps it is to confuse or mislead....or maybe there are other follow-on questions which need the infor given to get an answer.....but
xf = xi + vt + 1/2 at^2 xi=initial position......we will call this ZERO
xf=vt + 1/2 at^2 Problem now is that we do not know 't'
Here is an equation to find t
vf = vi + at vi = velocity initial = 50 km/hr = 50 x1000/3600 = 13.888 m/s vf will be zero when the car is stopped
0 = 13.888 m/s + (-6.6 m/s^2)(t) yields t= 2.104 seconds to come to a stop ....sub this into the first equation
xf = 13.888 m/s (2.104 s) + 1/2(-6.6 m/s^2)(2.104 s)^2 yields xf= 29.22 +(-14.61) = 14.61 m
