I need some help with these 2 problems please!! Two trains, a freight train and

Question

I need some help with these 2 problems please!!

Two trains, a freight train and a commuter train, are traveling in the same direction on parallel tracks. At t = 0 s, the front of the commuter train is some distance ahead of the front of the freight train. Be consistent, and measure everything from the front of each train. The graph shows the velocities of the two trains from t = 0 s to t = 60 s. Note that both trains have a constant acceleration until t_1 = 24.0 s, and then they both travel at constant velocity. Take v = 4.00 m/s. (a) Although the freight train starts behind the commuter train, it eventually catches up. Specifically, there is one instant when the two trains are exactly side-by side. However, after this instant, the commuter train gets ahead again-the freight train is never ahead. At what instant are the two trains side-by-side? t = 12 s (b) What is the distance between the two trains at t = 0 s? 124 m (c) At the end of the 60-second period, how far is the commuter train ahead of the freight train? 144 m As of September, 2017, NASA's Dawn spacecraft is nearing the end of its decade-long mission. You can read more about it here. It orbited two protoplanets (Vesta and Ceres) in the asteroid belt, between Mars and Jupiter, sending back plenty of excellent photos. The Dawn spacecraft was the first spacecraft to rely on ion propulsion. Ions are accelerated by high voltage to high speeds, propelling the spacecraft in the opposite direction. The acceleration can be rather small. Let's say that the spacecraft accelerates from rest to 50 miles per hour over a time period of 5 days. Assuming the acceleration is constant, what is the magnitude of the acceleration? 5.15e-5 m/s/s (b) Despite this small acceleration, fairly high speeds can be reached, and significant distance covered, by having the spacecraft accelerate for a long time. How far would Dawn travel, starting from rest, with the acceleration above taking place for 700 days? 3.13e3 m

Explanation / Answer

Given that

initial velocity u=0 m/s

finitial velocity v=50 mph=22.4 m/s

time t=5days =432000 sec

now we find the acceleration

acceleration a=22.4-0/432000=5.19*10^-5 m/s^2

now we find the distance at time t=700 days=6.05*10^8 sec

distance s=1/2at^2=1/2*5.19*10^-5*(6.05*10^8)^2=94.98*10^11 m

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