Question

A train starts from rest at station A and accelerates at 0.6 m/s^2 for 60 s. Afterwards it travels with a constant velocity for 25 min. It then decelerates at 1.2 m/s^2 until it is brought to rest at station B. Determine the distance between the stations.

Answer 1

Answer:

The distance between the station A and B will be:

$x_{A-B}=55.620\: km$  

Explanation:

Let's find the distance that the train traveled during 60 seconds.

$x_{1}=x_{0}+v_{0}t+0.5at^{2}$

We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:

$x_{1}=\frac{1}{2}(0.6)(60)^{2}$

$x_{1}=1080\: m$

Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.

$v_{1}=v_{0}+at$

$v_{1}=(0.6)(60)=36\: m/s$

Then the second distance will be:

$x_{2}=v_{1}*1500$

$x_{2}=(36)(1500)=54000\: m$        

The final distance is calculated whit the decelerate value:

$v_{f}^{2}=v_{1}^{2}-2ax_{3}$

The final velocity is zero because it rests at station B. The initial velocity will be v(1).

$0=36^{2}-2(1.2)x_{3}$

$x_{3}=\frac{36^{2}}{2(1.2)}$  

$x_{3}=540\: m$

Therefore, the distance between the station A and B will be:

$x_{A-B}=x_{1}+x_{2}+x_{3}$  

$x_{A-B}=1080+54000+540=55.620\: km$  

I hope it helps you!