Can someone explain to me how to solve this problem? A subway train starts from rest at a station and accelerates at a rate of 1.54m/s2 for 13.8s. It runs at constant speed for 70.9s and slows down at a rate of 3.45m/s2 until it stops at the next station. Find the total distance covered. Express your final answer in kilometers.
Break the problem down into three steps. Find the displacement during the first phase, while the train is accelerating for 13.8s. Then, find the displacement for the next 70.9s, while speed is constant. Repeat for the third phase when the train is slowing down. Add the three answers you get from each phase of the trip.
So would it be this for the first phase? x=1/2 at^2 x=1/2(1.54m/s^2)(13.8s)^2 x=146.6388m How do I get the second phase's acceleration at constant speed? x=1/2(___m/s^2)(70.9s)^2 And is the time still the same for the third phase? x=1/2(3.45m/s^2)(____s)^2
During the second phase there is no acceleration. The final velocity reached at the end of the first phase should be determined. Use this as the constant velocity during the second phase to find the distance travelled.
I still don't get it...I just know the answer is 1.78 km ):
for the first part we can use the given conditions and equation of motion to find distance S= ut + 1/2 a t^2 since train starts from rest means initial velocity is =0 s=1/2 *1.54*13.8^2 s=146.64m after 13.8 seconds we know it runs at a constant speed, we need to find that constant speed. acceleration is actually change in velocity divided by time. it started from rest and accelerated at 1.54ms^2 v-u/t=a v-0/13.8=1.54 v= 21.252 it runs at constant speed for 70.9 seconds . distance = speed * time 21.252 * 70.9 = 1506.8m last part in which the train decelerates we need to find the time in which it stops and we will find it using the relation v-u/t=a 0-21.252/t=-3.45 t=6.16s distance = speed * time 21.252 * 6.16 = 130.91m add all the distances = 1783.55 divide by 1000 to convert in Km = 1.78Km
Thank you so much for the explanation! It's been bothering me all week!
no problem this is what this community is all about. good luck :)
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