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.

As a train accelerates uniformly it passes

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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

 

Final Answers:

v=14 m/s

t=250 s

 

This question can be found in Engineering Mechanics: Dynamics (SI edition), 13th edition, chapter 12, question 12-30.

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