In one study of automobile stopping distances, the Acura, going at 70 mph, was a
ID: 2861496 • Letter: I
Question
In one study of automobile stopping distances, the Acura, going at 70 mph, was able to stop in the shortest distance. The brakes were applied so that there was a constant negative acceleration e applied so that there was a constant negative acceleration of 33.56 f /sec Show how to determine how long it will take the car to come to a stop. (Give answer with as many decimal places as your calculator can show.) a) b) How far will the car travel from the time the brakes are applied until it stops? (Give answer with as many decimal places as your calculator can show.)Explanation / Answer
a) given initial velocity u = 70mph = (70/3600)miles/sec,
acceleration a = -33.56ft/sec2 = -(33.56*0.000189394)miles/sec2
final velocity v= 0 (as automobile will stop finaly)
a = dv/dt
hence -(33.56*0.000189394)miles/sec2 = 0-(70/3600)/t miles/sec2 ...................(where t is stopage time)
hence t = 70/(3600*33.56*0.000189394) sec
t= 3.05919647835 seconds (automobile will stop after t seconds after driver applies break)
(b)now using 3r equation of motion i.e 2as = v2 - u2 , where s is the distance travelled by a vechile moving with acceleration a and inital , final velocities as u and v respectively
2*(-(33.56*0.000189394)miles/sec2)*s = 02 - ((70/3600)miles/se)2
hence s = (70/3600)2/(2*33.56*0.000189394) miles
= 0.02975067049 miles
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