a) An 1100-kg airplane starts from rest; 8.0s later reachesits takeoff speed of
ID: 1668721 • Letter: A
Question
a) An 1100-kg airplane starts from rest; 8.0s later reachesits takeoff speed of 35m/s. What is the average acceleration of theairplane during this time? b) A car is speeding up and has an instantaneous velocity of1.0 m/s in the +x-direction when a stopwatch reads 10.0s. It has aconstant acceleration of 2.0m/s2 in the +x-direction. 1)what change in speed occurs between t= 10.0s and t=12.0s? 2) Whatis the speed when the stopwatch reads 12.0s? c) A train is traveling along a straight, level track at 26.8m/s (60mi/h). Suddenly the engineer sees a truck stalled on thetracks 184 m ahead. If the maximum possible braking accelerationhas magnitude 1.52 m/s2, can the train be stopped intime... show work .. a) An 1100-kg airplane starts from rest; 8.0s later reachesits takeoff speed of 35m/s. What is the average acceleration of theairplane during this time? b) A car is speeding up and has an instantaneous velocity of1.0 m/s in the +x-direction when a stopwatch reads 10.0s. It has aconstant acceleration of 2.0m/s2 in the +x-direction. 1)what change in speed occurs between t= 10.0s and t=12.0s? 2) Whatis the speed when the stopwatch reads 12.0s? c) A train is traveling along a straight, level track at 26.8m/s (60mi/h). Suddenly the engineer sees a truck stalled on thetracks 184 m ahead. If the maximum possible braking accelerationhas magnitude 1.52 m/s2, can the train be stopped intime... show work ..Explanation / Answer
a) average acceleration = v / t = 35 - 0 / 8 - 0 = 35 / 8 = 4.375 m/s2 b) change in speed = at = 2. * (12 - 10) = 4 m/s c) speed = at = 2.0 * 12 = 24 m/s c) the time taken by train to travel a distance 184 m is S = ut + 0.5at2 ==> 184 = 26.8 t +0.5 * -1.52 * t2 ==> 0.76t2 - 26.8t + 184 =0 ==> t2 - 37.63 t + 242.1 = 0 Solving this we get t =5.6s
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