First, let's calculate the distance traveled during the acceleration phase:
Using the formula:
$v = u + at$
where:
v = final velocity (constant speed)
u = initial velocity (0 m/s)
a = acceleration (4.0 m/s^2)
t = time (6.0 s)
We can calculate the final velocity:
$v = 0 + (4.0)(6.0) = 24.0 m/s$
Next, we calculate the distance traveled during acceleration using the formula:
$s = ut + 0.5at^2$
$s = (0)(6.0) + 0.5(4.0)(6.0)^2$
$s = 0 + 0.5(4.0)(36.0)$
$s = 0 + 0.5(144.0)$
$s = 0 + 72.0$
$s = 72.0 m$
Now, let's calculate the distance traveled during the constant speed phase:
Distance = 3000m - 72m = 2928m
Now, we calculate the time it takes for the train to pass the checkpoint during the constant speed phase:
Time = Distance / Speed = 2928m / 24.0 m/s = 122s
Adding the time taken for acceleration:
Total time = 6.0s + 122s = 128s
Therefore, it takes 128 seconds for the train to pass the 3.0km checkpoint.
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.
If the next station is 3.0km from Clareview station, calculate the time it takes for this train to pass by this check point.
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