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:

Show me the final answer↓

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. 

 

This question can be found in Engineering Mechanics: Dynamics (SI edition), 13th edition, chapter 12, question 12-15.

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