# As a train accelerates uniformly it passes

As a train accelerates uniformly it passes successive kilometer marks while traveling at velocities of 2 m/s and then 10 m/s. Determine the train’s velocity when it passes the next kilometer mark and the time it takes to travel the 2-km distance.

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#### Solution:

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Let us figure out the acceleration of the train. To do so, we will use the following kinematics 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)

Let us substitute the values we know:

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

$10^2=2^2+2a(1000-0)$

(Here, the train travels 1000 m, starting at the origin of $s_0=0$ m. At the origin, it’s speed is 2 m/s, and at the 1000 m mark, the train has a speed of 10 m/s)

$a=0.048$ m/$s^2$

We can now find the speed of the train at the next marker. Note that the initial velocity of the train is now 10 m/s and we are trying to find the final velocity. As before, the distance between the markers is 1000 m. Using the same kinematics equation, we can write:

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

$v^2=10^2+2(0.048)(1000-0)$

$v=14$ m/s

We can calculate the time it took for the train to reach the next marker 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)

Substitute the values we found:

$v=v_0+at$

$14=2+(0.048)(t)$

(Remember, we are looking for the time for the full journey. That means the initial velocity of the train was 2 m/s and final velocity was 14 m/s)

$t=250$ s

$v=14$ m/s
$t=250$ s