A express train travels along a straight line between two stations, A and D. Stations B and C are spaced evenly along the line between A and D. The engineer leaves A and accelerates at a constant rate until he passes station B, at which time he coasts with constant velocity. When he reaches station C, he slows down at the same rate he used for the first segment of the trip, thereby coming gently to rest at station D. Given that the total trip requires 5 mins, how much time is spent going from
1) A to B?
2) B to C?
3) C to D?
Please help me ASAP!!
Let's say that the three times to traverse AB,BC,CD are x,y,z respectively. They add to 5 min.
The time to accelerate is the same as the time to decelerate, so x=z.
If the acceleration is a m/s^2, we have
x+y+z = 5
1/2 ax^2 = axy
1/2 ax^2 = axz - 1/2 az^2
We can factor out the a from all the terms, and we are left with
2x+y = 5
x=2y
Now solve for x and y
Which physics equations should I use?
To solve this problem, we can use the equations of motion to determine the time spent traveling between each station. Let's break down each segment of the trip.
1) A to B:
Let's assume the distance AB is 'd' and the acceleration during this segment is 'a'.
Using the equation of motion, we have:
d = (1/2) * a * t^2, where t is the time taken to travel from A to B.
We also know that the total time for the trip is 5 minutes.
So, t + (5 - t)/2 = t + 2.5 - 0.5t = 2.5 minutes.
Solving the equation, we get t = 2 minutes.
Therefore, the time spent going from A to B is 2 minutes.
2) B to C:
In this segment, the train coasts with a constant velocity, meaning there is no acceleration involved. Therefore, the time taken to travel from B to C is the same as the time taken to travel from A to B, which is 2 minutes.
3) C to D:
In this segment, the engineer applies the same deceleration as in segment A to bring the train gently to rest at station D.
Since the train is coming to rest, its final velocity is 0.
Using the equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
We already know the initial velocity u, which is the same as the velocity when the train reaches station C.
We also know the acceleration a, which is the same as in segment A.
Now, we need to find the time t.
Since the time taken to travel from B to C is also 2 minutes, and the total trip time is 5 minutes, the time taken to travel from C to D can be found by subtracting these two times from the total trip time:
t + 2 + 2 = 5 minutes.
Solving this equation, we get t = 1 minute.
Therefore, the time spent going from C to D is 1 minute.
So, to summarize:
1) A to B: 2 minutes
2) B to C: 2 minutes
3) C to D: 1 minute
I hope this explanation helps!