a train starts from rest and accelerates unifotmly, until it has travled 3.7 km and acquired a velocity of 30 m/s. the train then moves at a constant velocity of 30 m/s for 410 s. the train then slows down uniformly at 0.065 m/s^2, until it reaches a halt. what distance does the train travel while it is slowing down
See previous post.
To find the distance the train travels while slowing down, we need to break down the problem into different segments and calculate the distance traveled in each segment.
Let's calculate the distance traveled during each segment:
1. Segment 1: Acceleration
The train starts from rest and accelerates uniformly until it reaches a velocity of 30 m/s. We need to find the distance covered during this time.
We know the initial velocity (u) is 0 m/s, final velocity (v) is 30 m/s, and acceleration (a) is constant. The formula to find the distance (s) in this segment is:
s = (v^2 - u^2) / (2a)
Substituting the given values:
s = (30^2 - 0^2) / (2a)
We still need to find "a," the acceleration, which is not given. However, we can calculate it using the formula:
a = (v - u) / t
Here, v = 30 m/s, u = 0 m/s, and we need to find "t" (the time taken to reach 30 m/s).
Using the formula:
t = (v - u) / a
= (30 - 0) / a
= 30 / a
Now, substituting this value of "t" back into the distance formula:
s = (30^2 - 0^2) / (2 * (30 / a))
= a
So, the distance covered during acceleration is equal to "a."
2. Segment 2: Constant Velocity
The train moves at a constant velocity of 30 m/s for 410 seconds. The formula to find the distance (s) in this segment is:
s = v * t
= 30 * 410
So, the distance covered during this time is 12,300 meters (or 12.3 km).
3. Segment 3: Deceleration
The train slows down uniformly at an acceleration of -0.065 m/s^2 until it comes to a halt. We need to find the distance covered during slowing down.
Here, the final velocity (v) is 0 m/s, and the acceleration (a) is -0.065 m/s^2.
Using the formula of motion:
v^2 = u^2 + 2a * s
Since the final velocity (v) is 0 m/s:
0 = 30^2 + 2(-0.065)s
Simplifying the equation:
900 - 0.13s = 0
-0.13s = -900
s = -900 / -0.13
So, the distance covered during deceleration is approximately 6,923.08 meters (or 6.92 km).
Therefore, the total distance traveled while slowing down is approximately 6,923.08 meters (or 6.92 km).