The LRT train in Edmonton starts from rest at the Clareview station and accelerates at a rate of 4.0 m/s^2 for the first 6.0s. The train then maintains a constant speed for the next 45.0s.
determine the displacement of the train after 51.0s of travel
To determine the displacement of the train after 51.0s of travel, we first need to calculate the distance covered during the acceleration phase and the constant speed phase separately.
1. Acceleration phase:
Using the equation for distance covered during constant acceleration:
d = 0.5 * a * t^2
where:
d = distance
a = acceleration
t = time
We can plug in the values:
d = 0.5 * 4.0 m/s^2 * (6.0s)^2
d = 0.5 * 4.0 * 36.0
d = 72.0 meters
2. Constant speed phase:
Since the speed is constant, the distance traveled is simply given by:
d = v * t
where:
d = distance
v = constant speed
t = time
Given that the constant speed is achieved after 6.0s and maintained for 45.0s, we have:
d = v * 45.0s
3. Total displacement:
Adding the distance covered during acceleration and the distance covered at constant speed, the total displacement can be calculated as:
Total displacement = 72.0m + v * 45.0s
Now, to calculate the constant speed (v), we consider the distance traveled during the acceleration phase as the initial velocity at the start of the constant speed phase. So, the initial velocity (u) after 6.0s of acceleration is given by:
u = a * t
u = 4.0 m/s^2 * 6.0s
u = 24.0 m/s
This becomes the constant speed (v) during the constant speed phase.
Hence, the total displacement after 51.0s of travel is:
Total displacement = 72.0m + 24.0 m/s * 45.0s
Total displacement = 1173.0 meters
Therefore, the displacement of the train after 51.0s of travel is 1173.0 meters.