A train starts from rest at station A and accelerates at 0.5 m/s^2 for 60 s. Afterwards it travels with a constant velocity for 15 min. It then decelerates at 1 m/s^2 until it is brought to rest at station B. Determine the distance between the stations.
Solution:
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Let us first figure out how far the train traveled while it is accelerating during the 60 seconds. We can use the following kinematics equation to figure it out.
(Where s is final displacement, s_0 is initial displacement, v_0 is initial velocity, t is time, and a is constant acceleration)
Let us substitute the values we know:
s=0+0+\dfrac{1}{2}(0.5)(60^2)
(Here, the initial velocity is 0 m/s since the train starts from rest, and the initial displacement is zero since that will be our origin)
s=900 m
We now need to find the velocity of the train after accelerating. It can be found using the following kinematics equation:
(Where v is final velocity, v_0 is initial velocity, a is acceleration and t is time)
v=0+(0.5)(60)
v=30 m/s
Since the train traveled at constant velocity for 15 mins (900 seconds), we can figure out the distance traveled using the equation we used before.
s=0+(30)(900)+0
(Here, our initial displacement is 0 because that will be our origin, the acceleration is also 0 because the train is traveling at constant velocity)
s=27000 m
Let us now calculate the distance traveled by the train while it decelerated. We can use the following kinematic equation:
(Where v is final velocity, v_0 is initial velocity, a is constant acceleration, s is final displacement, and s_0 is initial displacement)
Substitute the values we know:
(Here, our final velocity is 0 m/s because the train comes to a stop. The acceleration is negative because the train is slowing down, and our initial velocity is 30 m/s since we found the value in the previous step. Also note that our initial displacement is 0 m since it is our origin)
s=450 m
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