Final answer:
The train accelerates with a uniform acceleration and reaches 18 m/s when the rear of the train reaches the end of the platform. Using the kinematic equations for uniformly accelerated motion, we find that the train's acceleration is 0.0257 m/s². The time taken for the train to reach 18 m/s from rest is then calculated to be 700 s.
Step-by-step explanation:
To solve this problem, we must first recognize that the acceleration of the train is constant, and when the rear of the train reaches the end of the platform, it is traveling at 18 m/s.
This is an application of the kinematic equations for uniformly accelerated motion. We can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the train starts from rest, u is 0, and when the rear of the train reaches the end of the platform, v is 18 m/s.
The next step involves finding the acceleration.
To do this, we can use the information when the front of the train reaches the end of the platform, traveling at 3 m/s. Assuming the front of the train and the platform edge are the starting and ending points, the train covers its own length, 175 m, during acceleration.
We use the equation v² = u² + 2as, where s (displacement) is the length of the train (175 m), u is the initial velocity (0 m/s), and v is the final velocity (3 m/s).
Substituting the values, we have 3² = 0² + 2a*175, which simplifies to 9 = 350a.
Solving for a gives a = 9/350 = 0.0257 m/s².
Finally, using the acceleration and initial velocity of 0, we can find the time taken to reach 18 m/s with t = (v - u) / a. Therefore, t = (18 m/s - 0 m/s) / 0.0257 m/s², which gives us t = 700 s.