A train going from one station to the next on a straight section of track accelerates from rest at 1.4m/s^2 for 15s. It then proceeds at constant speed for 1100m before slowing down at 2.2m/s^2 until it stops at the station. What is the distance between the stations? How long does it take the train to go between stations?

Question

A train going from one station to the next on a straight section of track accelerates from rest at 1.4m/s^2 for 15s. It then proceeds at constant speed for 1100m before slowing down at 2.2m/s^2 until it stops at the station. What is the distance between the stations? How long does it take the train to go between stations?

Answer 1

figure the distances during the acceleration, and the final velocity.

Then figure out the distance to deaccelerate.

Answer 2

o...i got that, but what formula do u use to find the time it takes to go between stations?

Answer 3

To find the distance between the stations, we can break down the train's motion into three parts: the initial acceleration, the constant speed, and the final deceleration.

1. Initial acceleration:
Given: Initial velocity (u) = 0 m/s, Acceleration (a1) = 1.4 m/s^2, Time (t1) = 15 s

To find the distance covered during this time, we can use the formula:

Distance (s1) = u * t1 + (1/2) * a1 * t1^2

Substituting the given values, we have:

s1 = 0 * 15 + (1/2) * 1.4 * (15^2)
s1 = 0 + (1/2) * 1.4 * 225
s1 = 157.5 meters

2. Constant speed:
The train travels at a constant speed and covers a distance of 1100 meters.

3. Final deceleration:
Given: Final velocity (v) = 0 m/s, Deceleration (a2) = 2.2 m/s^2

We can use the formula for deceleration:

Final velocity (v) = Initial velocity (u) + a2 * t2

Since the final velocity is 0 m/s and the initial velocity is the constant speed, we can write:

0 = Constant speed + 2.2 * t2

Solving for t2, we get:

t2 = -Constant speed / 2.2

Combining these three distances, the total distance between the stations is:

Total distance = s1 + Constant speed distance + s3
Total distance = 157.5 + 1100 + s3

To find the value of s3, we can use the formula for deceleration:

s3 = (Constant speed)^2 / (2 * a2)

Substituting the given values, we have:

s3 = (Constant speed)^2 / (2 * 2.2)
s3 = (Constant speed)^2 / 4.4

Now, the total distance becomes:

Total distance = 157.5 + 1100 + (Constant speed)^2 / 4.4

To find the total time taken, we add up the times for each segment:

Total time = t1 + Constant speed time + t2
Total time = 15 + (Constant speed / Constant speed) + (-Constant speed / 2.2)

We can simplify this further:

Total time = 15 + 1 + (-Constant speed / 2.2)
Total time = 16 - (Constant speed / 2.2)

To solve for the distance and time, we need the value of the constant speed. Please provide that value.

figure the distances during the acceleration, and the final velocity.

o...i got that, but what formula do u use to find the time it takes to go between stations?

LOCAL

To find the distance between the stations, we can break down the train's motion into three parts: the initial acceleration, the constant speed, and the final deceleration.