A car starts from rest and accelerates at a(t) = 2 + 2.4 t (ft/s2) for 5 [s], th

Question

A car starts from rest and accelerates at a(t) = 2 + 2.4 t (ft/s2) for 5 [s], then, it continues with a constant acceleration a(t) = 16 [ft/sec2] for another 10 [s]; at this point the driver applies the brakes with the car moving at a constant acceleration of - 10 [ft/sec2], until it comes to rest. Determine: The maximum velocity of the car and the time at which this velocity is reached The total distance traveled by the car. The total time of travel Graph the acceleration, velocity and displacement of the car from t=0 through t = ttot Identify the curves with the corresponding equations,

Explanation / Answer

a)dv/dt=a,,,,,,,,, dv=adt,,,,,,,, integrate,,,,,,,, dv=[2+2.4t]dt,,,,,,,,,, v=2t+1.2t^2,,,,,,,,,from 0 to 5,,,,,,,, v=40 ft/s,,,,,,,,,,,, for next 10 sec,,,,,,,,,,,, V=v+16t,,,,,,,,,,, V=40+16x10=200 ft/s=maximum velocity. at total time=10+5=15 seconds,,,,,,,,,,,, c) then car deaccelerates,,,,,,at a= -10 ft/s 0=V-10t,,,,,,,,,, 200=10t,,,,,,,,,, t=20 sec,,,,,,,,,,, total time of travel=5+10+20=35 seconds ,,,,,,,,,,,,,,,, b) total distance=S1+S2+S3=[2t+1.2t^2]dt + 40t+at^2/2+2000=75+1200+2000=3275 ft

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