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
a) A to B?
b) B to C?
c) C to D?
To solve this problem, we can use the formulas of motion under constant acceleration. Let's assume that the distance between A and D is represented by "d".
a) To find the time spent going from A to B, we first need to determine the equations governing the motion during this segment of the trip.
Let's denote:
t1 = time taken from A to B
v1 = constant velocity reached at B
a1 = acceleration during A to B segment
Since the engineer accelerates at a constant rate until he passes station B, we can use the following equation of motion:
v = u + at
In this case, the initial velocity (u) is 0, as the engineer starts from rest. The final velocity (v) is v1, and the time (t) is t1. The acceleration (a) is constant and is given by a1. Therefore, the equation becomes:
v1 = 0 + a1 * t1 Equation 1
We are also given that the engineer coasts with constant velocity from B to C. This means that there is no acceleration (a2 = 0) during this segment. Therefore, the equation becomes:
v2 = v1 Equation 2
Using Equation 2, we know that the velocity at C (v2) is equal to v1.
b) To find the time spent going from B to C, we need to know the distance between B and C. Since B and C are evenly spaced, we can assume that the distance between them is equal to "d/2".
Using the equation of motion mentioned earlier, we have:
v = u + at
In this case, the initial velocity (u) is v1 and the final velocity (v) is v1. Since there is no acceleration during this segment (a2 = 0), the equation becomes:
v1 = v1 + 0 * t2
Therefore, the time taken from B to C (t2) is 0. This means that the engineer spends no time traversing this segment.
c) To find the time spent going from C to D, we can use the equations of motion again.
Let's denote:
t3 = time taken from C to D
v3 = constant velocity reached at C
a3 = deceleration during C to D segment
Since the engineer slows down at the same rate he used for the first segment of the trip, the deceleration (a3) is equal to the acceleration (a1) during the A to B segment. Therefore, the equation becomes:
v3 = v2 + a3 * t3 Equation 3
We are also given that the engineer comes to rest gently at station D. This means that the final velocity (v3) is 0. Therefore, Equation 3 becomes:
0 = v2 + a3 * t3
Substituting Equation 2 (v2 = v1) into the above equation, we get:
0 = v1 + a3 * t3 Equation 4
We know that the total trip requires 5 minutes, so we can write:
t1 + t2 + t3 = 5 Equation 5
We can solve Equations 1, 4, and 5 simultaneously to find the values of t1, t2, and t3.
This is the approach to solve the problem. By substituting the values for acceleration and distance, you can calculate the time taken for each segment of the trip.