1. A train starts from rest andaccelerates uniformly, until it has traveled 2.1
ID: 1760782 • Letter: 1
Question
1. A train starts from rest andaccelerates uniformly, until it has traveled 2.1 km and acquired avelocity of 24 m/s. The train then moves at a constantvelocity of 24 m/s for 400 s. The train then slows downuniformly at 0.065 m/s2, until it is brought to ahalt.
a. (5 points) How longdoes it take the train to accelerate to 24 m/s?
b. (5 points) What is thedistance traveled by the train while slowing down?
c. (5 points) What is thetotal distance traveled by the train?
d. (5 points) What is theaverage speed of the train from the time it starts until it isbrought to a halt?
Explanation / Answer
a ) as the initial velocity of the train is zero U = 0 we have V = at and S =1/2 at2 t = 2S / V = 175 s the time taken to accelerate is t = 175 s b ) we have a = 0.065 m/s2 so S = V2 / 2a = 4430.76 m = 4.430 km c) the distance travelled with constant velocity is = V * 400s = 9600 m = 9.600 km so total distance travelled is = 2.1km +9.6km+4.43km = 16.13km d ) average speed = total distance travelled / totaltime taken time taken during deceleration is t = V / a = 369.23 s so the time T = 944.23 s = 0.26 h so the speed = 61.49 km / h so S = V2 / 2a = 4430.76 m = 4.430 km c) the distance travelled with constant velocity is = V * 400s = 9600 m = 9.600 km so total distance travelled is = 2.1km +9.6km+4.43km = 16.13km d ) average speed = total distance travelled / totaltime taken time taken during deceleration is t = V / a = 369.23 s so the time T = 944.23 s = 0.26 h so the speed = 61.49 km / hRelated Questions
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