# A train starts from rest at station 1

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:

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.

$s=s_0+v_0t+\dfrac{1}{2}at^2$

(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:

$s=s_0+v_0t+\dfrac{1}{2}at^2$

$s=0+0+\dfrac{1}{2}(0.5)(60^2)$

(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:

$v=v_0+at$

(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.

$s=s_0+v_0t+\dfrac{1}{2}at^2$

$s=0+(30)(900)+0$

(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:

$v^2=v_0^2+2a(s-s_0)$

(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:

$0=30^2+2(-1)(s-0)$

(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

Thus, the total distance traveled (distance between the two stations) is 900 + 27000 + 450 = 28350 m = 28.35 km.