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
All that stuff at the beginning is irrelevant here. When the train starts to slow down,
v=30m/s
a = -.065 m/s^2
So, it takes 30/.065 = 461.5 seconds to stop
s = 30t - .0325t^2 = 6923m = 6.9 km
. A train starts from rest and accelerates uniformly, until it has traveled 3.7 km and acquired a velocity of The train then moves at a constant velocity of for 410 s. The train then slows down uniformly at until it reaches a halt. What distance does the train travel while it is slowing down?
To solve this problem, we can break it down into three parts: the acceleration phase, the constant velocity phase, and the deceleration phase.
1. Acceleration Phase:
Given:
Initial velocity (u) = 0 m/s
Final velocity (v) = 30 m/s
Distance (s) = 3.7 km = 3700 m
We can use the formula for distance with uniform acceleration:
s = (v^2 - u^2) / (2a)
By rearranging the formula, we can solve for acceleration (a):
a = (v^2 - u^2) / (2s)
a = (30^2 - 0^2) / (2 * 3700)
a = 900 / 7400
a = 0.1216 m/s^2
2. Constant Velocity Phase:
The train moves at a constant velocity of 30 m/s for 410 s. The distance covered in this phase can be calculated by multiplying the velocity by time:
s = v * t
s = 30 * 410
s = 12,300 m
3. Deceleration Phase:
Given:
Deceleration (a) = -0.065 m/s^2 (negative because it is slowing down)
Initial velocity (u) = 30 m/s
Final velocity (v) = 0 m/s
We can use the formula for distance with uniform acceleration again to find the distance traveled during the deceleration phase:
s = (v^2 - u^2) / (2a)
By rearranging the formula, we can solve for distance (s):
s = (v^2 - u^2) / (2a)
s = (0^2 - 30^2) / (2 * -0.065)
s = (0 - 900) / -0.13
s = -900 / -0.13
s = 6923.08 m (rounded to 2 decimal places)
Therefore, the train traveled approximately 6923.08 meters while slowing down.
To find the distance the train travels while slowing down, we can break down the problem into three parts: the acceleration phase, the constant velocity phase, and the deceleration phase.
1. Acceleration phase:
We know that the train starts from rest and accelerates uniformly until it reaches a velocity of 30 m/s. We are given this final velocity, so we can use the formula of acceleration to calculate the acceleration (a) during this phase.
The formula for acceleration (a) is:
v = u + at
where:
v = final velocity = 30 m/s
u = initial velocity = 0 m/s (as the train starts from rest)
t = time taken to reach the final velocity
Rearranging the formula, we get:
a = (v - u) / t
Substituting the given values:
a = (30 m/s - 0 m/s) / t
We don't have the time (t) given for this phase, so we need to find it. We can use another formula to do that.
The formula for displacement (s) during uniform acceleration is:
s = ut + (1/2)at^2
Using this formula with the given final displacement of 3.7 km (which is equal to 3700 m) and initial velocity (u) of 0 m/s, we have:
3700 m = 0 m/s * t + (1/2)a * t^2
Simplifying this equation, we get:
(1/2)a * t^2 = 3700 m
Now we can substitute the value of "a" from the first formula, which is (30 m/s - 0 m/s) / t, into the second formula:
(1/2) * ((30 m/s - 0 m/s) / t) * t^2 = 3700 m
Simplifying further, we get:
(1/2) * 30 * t = 3700 m
15t = 3700 m
t = 3700 m / 15
t = 246.67 s
Now that we have the value of t, we can substitute it back into the first formula to find the acceleration:
a = (30 m/s - 0 m/s) / t
a = 30 m/s / 246.67 s
a ≈ 0.1216 m/s^2
2. Constant velocity phase:
During this phase, the train moves at a constant velocity of 30 m/s for 410 s. Since we are given the constant velocity and the time, we can directly use the formula for distance (s) for constant velocity motion, which is:
s = v * t
Substituting the given values:
s = 30 m/s * 410 s
s = 12300 m
3. Deceleration phase:
In this phase, the train slows down uniformly at an acceleration of -0.065 m/s^2 (negative because it is in the opposite direction to the motion). We need to find the distance traveled during this phase.
We can use the formula for distance (s) during uniform deceleration, which is the same as the formula for uniform acceleration:
s = ut + (1/2)at^2
In this case, the final velocity (v) is 0 m/s (the train comes to a halt), the initial velocity (u) is 30 m/s (the velocity it had during the constant velocity phase), and the acceleration (a) is -0.065 m/s^2. We need to find the time taken (t) during this phase.
The formula becomes:
0 m = 30 m/s * t + (1/2)(-0.065 m/s^2) * t^2
Simplifying this equation, we get:
15t - 0.0325t^2 = 0
Now we can solve this quadratic equation to find the values of t. Factoring out t, we have:
t(15 - 0.0325t) = 0
This equation gives us two possible solutions: t = 0 (which is not relevant in this case) and t = 461.54 s.
Since we're interested in the time during the deceleration phase, we can ignore the t = 0 solution. Therefore, t = 461.54 s.
Now that we have the value of t, we can substitute it back into the formula for distance to find the distance traveled during this phase:
s = 30 m/s * 461.54 s + (1/2)(-0.065 m/s^2) * (461.54 s)^2
Simplifying the equation, we get:
s ≈ 6911.04 m
Therefore, the distance the train travels while slowing down is approximately 6911.04 meters.