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
To find the distance the train travels while slowing down, we need to break down the problem and calculate the distances during each phase of the train's motion: acceleration, constant velocity, and deceleration.
1. Acceleration phase:
Given that the train starts from rest and acquires a velocity of 30 m/s while traveling 3.7 km, we can use the following equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (30 m/s)
u = initial velocity (0 m/s, since the train starts from rest)
a = acceleration
s = distance
We need to find the acceleration (a) during the acceleration phase. Rearrange the equation:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (30^2 - 0^2) / (2 * 3.7 km)
Convert the distance to meters by multiplying it by 1000 since 1 km = 1000 m:
a = (30^2 - 0^2) / (2 * 3.7 km * 1000 m/km)
Simplifying:
a = (900 - 0) / (7.4 km * 1000 m/km)
a = 900 / 7.4 km * 1000
a ≈ 121.62 m/s^2
Now that we have the acceleration, we can calculate the distance traveled during this phase (s1) using the formula:
s1 = (v^2 - u^2) / (2a)
Substituting the given values:
s1 = (30^2 - 0^2) / (2 * 121.62 m/s^2)
s1 = 900 / 243.24 m/s^2
s1 ≈ 3.7 km
2. Constant velocity phase:
The train moves at a constant velocity of 30 m/s for 410 s. The distance traveled during constant velocity is found using:
s2 = v × t
Substituting the given values:
s2 = 30 m/s × 410 s
s2 = 12300 m
3. Deceleration phase:
The train slows down uniformly at a rate of 0.065 m/s^2 until it comes to a halt. We need to find the distance (s3) traveled during this phase.
Using the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s)
u = initial velocity (30 m/s)
a = acceleration (-0.065 m/s^2, negative because it is decelerating)
s = distance
Rearranging the equation:
s = (v^2 - u^2) / (2a)
Substituting the given values:
s3 = (0^2 - 30^2) / (2 * (-0.065 m/s^2))
s3 = (-900) / (-0.13 m/s^2)
s3 ≈ 6923.08 m
Therefore, the distance traveled by the train while slowing down is approximately 6,923.08 meters (or 6.92308 km).