Answer: Total distance = 28350 m
Explanation: We can divide the distance travel in 3 steps: one accelerating, the middle step at constant velocity and a final one decelerating.
For the first distance (x1) , we know the acceleration from rest (0.5 m/sec²) and the time traveled (60 sec).
By definition, acceleration is equal to the change in velocity over time, so we can put the following:
a= vf-vo/Δt⇒ if vo=0 (starts from rest) ⇒vf= a*Δt=0.5 m/sec²*60 sec=30 m/sec.
Starting from rest, we can write that x1= 1/2 * a* Δt²= 900 m.
(we arrive to the same result applying vf²=2*a*Δx).
For the second part, we know that at t=60 sec, v= 30 m/sec.
As the second part is traveled at constant velocity, by definition, we can write the following:
v= Δx/Δt (Δt=15 min= 15*60 sec/min=900 sec) ⇒Δx=v*Δt=30m/sec*900sec=27000 m
FInally, as the train is brought to rest after decelerating, and we know that it's decelerating at a rate of 1 m/sec2, and the initial velocity is 30 m/sec, we can get the time traveled while it was decelerating:
vf-vo= a*Δt ⇒ Δt=(vf-vo)/a ⇒Δt= 30 sec 9 (as vf=0)
With this information, we can use the following kinematic equation:
Δx= vo*t + 1/2 * a* t²
Replacing by the values that we have got previously, we got :
Δx= 900 m - 450 m = 450 m
Adding the three distances, we have:
Δx= 900 m + 27000 m + 450 m = 28350 m.
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