Answer to Question #15474 in Mechanics | Relativity for Siti zahidah
T = 72[km/h]/5[m/s²] = 0.7[s] + 72*1000/3600[m/s]/5[m/s²] = 4 s
to come to a complete stop after applying the brakes.
L = (0.7[s] + 4[s]/2)*72*1000/3600[m/s] ≈ 54 m
before coming to a stop. So, he'll stop 55[m] - 54[m] = 1 meter before the stop line.
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The path is
L = v*t0 + a*t^2/2
where v = 72 kmph is the initial velocity, a = 5 m/s^2 is the deceleration, t is time of decelerated motion and t0 = 0.7 s is a reaction time. We know that a*t = v, thus
L = v*t0 + v*t/2 = (t0 + t/2)*v
why must the t=4s divide 2 ???