A freight train has a mass of 2.6 × 10^7 kg. If a locomotive can exert a constant pull of 5.5 × 10^5 N, how long would it take to increase the speed of the train from rest to 89.4 km/h.

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ndnordic's profile pic

ndnordic | High School Teacher | (Level 2) Associate Educator

Posted on

This can be solved by breaking the problem into two parts.

You know the mass and force applied to the locomotive, so you can calculate the acceleration of the locomotive starting from rest.

F = ma, so a = F/m

5.5 x 10^5 N/2.6 x 10^7 kg = 0.02115 m/s/s

Now the train will acceleration from rest, to a final speed of 89.4 km/hr so you need to convert that to m/s.

89.4 km/hr * 1000 m/km* 1hr/3600 s = 24.83 m/s.

Using the kinematic equation which states:  Vf = Vi + at, where Vf is the final velocity, Vi is the initial velocity, a is the acceleration in m/s/s and t is the time in seconds, solve for t.

Vf = 24.83 m/s

Vi = 0

a = 0.02115 m/s/s

24.83 = 0 + 0.02115t

t = 1174.15 seconds = 19.57 min = 0.326 hours


justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The freight train has a mass of 2.6*10^7 kg. It is pulled by a locomotive that can exert a constant force of 5.5*10^5 N.

The constant force accelerates the freight train. Force is equal to the product of mass and acceleration. Modifying the formula acceleration a = force/mass.

Substituting the values we have here, a = 5.5*10^5/2.6*10^7 = 0.021 m/s^2.

If an object at a velocity u is accelerated by a, the velocity after t seconds is v = u + a*t.

Here, we need to increase the velocity of the train from rest to 89.4 km/h. 89.4 km/h = 89.4*1000/3600 = 24.83 m/s

As the initial velocity is 0, we have 24.83 = 0.021*t

=> t = 1182.5 s

The time required to increase the speed of the train to 89.4 km/h is 1182.5 s

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