6 t - 16 = 1/2 1 t^2 ... 12 t - 32 = t^2 ... t^2 - 12 t + 32 = 0 ... (t - 8) (t - 4) = 0
looks like he catches the bus in four seconds
... if he kept running, he would pass the bus
... which would catch back up to him at the eight second mark
running for four seconds means he travels 24 m
a pedestrian is running at his maximum speed of 6.0 m/s trying to catch a bus that is stopped at a traffic light. when he is 16m from the bus, the light changes and the bus pulls away from the pedestrian with an acceleration of 1.0 m/s^2. does the pedestrian catch the bus and, if so, how far does he have to run. (if not, what is the pedestrian's distance of closest approach)
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