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 can break down the problem into three parts.
Part 1: Acceleration phase
In this part, the train starts from rest and accelerates uniformly until it reaches a velocity of 30 m/s. The distance traveled during any phase of constant acceleration can be calculated using the equation:
s = (v^2 - u^2) / (2a),
where:
s = distance traveled,
v = final velocity,
u = initial velocity,
a = acceleration.
In this case, the final velocity (v) is 30 m/s (given), the initial velocity (u) is 0 m/s (rest), and we need to calculate the acceleration (a).
Using the equation:
v = u + at,
where t is the time taken to reach the final velocity, we can rearrange it to solve for acceleration:
a = (v - u) / t.
Let's substitute the values into the equation:
a = (30 m/s - 0 m/s) / t.
Since the train starts from rest, we know that the time taken (t) to reach 30 m/s is the same as the time taken to acquire a distance of 3.7 km.
Converting 3.7 km to meters:
3.7 km = 3700 m.
Now we can solve for time (t):
t = s / v,
t = 3700 m / 30 m/s,
t ≈ 123.33 s.
Now we can calculate the acceleration (a):
a = (30 m/s - 0 m/s) / 123.33 s,
a ≈ 0.243 m/s^2.
Using the equation for distance (s = (v^2 - u^2) / (2a)), we can find the distance traveled during the acceleration phase:
s1 = (30 m/s)^2 - (0 m/s)^2 / (2 * 0.243 m/s^2),
s1 = 900 m / 0.486 m/s^2,
s1 ≈ 1851 m.
Therefore, during the acceleration phase, the train travels approximately 1851 meters.
Part 2: Constant velocity phase
During this phase, the train moves at a constant velocity of 30 m/s for 410 seconds. The distance traveled at constant velocity is given by the equation:
s = v * t,
where:
s = distance traveled,
v = velocity,
t = time.
Substituting the values:
s2 = 30 m/s * 410 s,
s2 = 12300 m.
Therefore, during the constant velocity phase, the train travels 12300 meters.
Part 3: Deceleration phase
In this phase, the train slows down uniformly at 0.065 m/s^2 until it comes to a halt (0 m/s).
Using the equation v = u + at, we can solve for the time taken (t) to decelerate:
0 m/s = 30 m/s - (0.065 m/s^2) * t,
30 m/s = 0.065 m/s^2 * t,
t = 30 m/s / 0.065 m/s^2,
t ≈ 461.54 s.
Using the equation for distance (s = (v^2 - u^2) / (2a)), we can find the distance traveled during the deceleration phase:
s3 = (0 m/s)^2 - (30 m/s)^2 / (2 * -0.065 m/s^2),
s3 = -900 m / -0.13 m/s^2,
s3 ≈ 6923 m.
Since distance cannot be negative, we will take the absolute value, so s3 ≈ 6923 m.
Therefore, during the deceleration phase, the train travels approximately 6923 meters.
Finally, we can calculate the total distance traveled by adding the distances from each phase:
Total distance = s1 + s2 + s3,
Total distance ≈ 1851 m + 12300 m + 6923 m,
Total distance ≈ 21074 m.
Therefore, while the train is slowing down, it travels approximately 21074 meters.