After getting T/369(I) speed over the points while passing the signal is ________ km/h. A. 75 B. 15 C. 25 D. 30

75
15
25
30

The correct answer is $\boxed{\text{B) 15 km/h}}$.

The formula for calculating the speed over the points while passing the signal is:

$$\text{Speed} = \frac{T}{369(I)}$$

where:

  • $T$ is the time in seconds it takes to travel from 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z"/> Subscribe on YouTube
the starting point to the signal,
  • $I$ is the grade of the track in percent, and
  • $369$ is a constant.
  • In this case, $T = 1$ second and $I = 1$ percent. Substituting these values into the formula, we get:

    $$\text{Speed} = \frac{1}{369(1)} = \boxed{15 \text{ km/h}}$$

    Option A is incorrect because it is the speed of the train after getting T/369(I) seconds. Option C is incorrect because it is the speed of the train after getting T/369(I) seconds and passing the signal. Option D is incorrect because it is the speed of the train after getting T/369(I) seconds and passing the signal on a level track.

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