A heavy freight train has a mass of 11800 metric tons. If the locomotive can pull with a force of 691000 N, how long does it take to increase it's speed from 0 to 70.3 km/h?
To find the time it takes for the train to increase its speed from 0 to 70.3 km/h, we can use Newton's second law of motion, which states that force equals mass multiplied by acceleration (F = m * a).
First, let's convert the mass of the train from metric tons to kilograms. Since 1 metric ton is equal to 1000 kilograms, the mass of the train is 11800 metric tons * 1000 kg/metric ton = 11,800,000 kg.
Next, let's convert the speed from km/h to m/s, as we are using the metric system in our calculations. To convert, we divide the speed (70.3 km/h) by 3.6, since there are 3.6 seconds in an hour. This gives us a speed of 70.3 km/h ÷ 3.6 = 19.5 m/s.
Now, we need to find the acceleration. Rearranging the formula F = m * a, we get a = F / m. Plugging in the values, we have a = 691000 N / 11,800,000 kg ≈ 0.0586 m/s^2.
Finally, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity (which is 0 m/s since the train starts from rest), a is the acceleration, and t is the time.
Plugging in the values, we have 19.5 m/s = 0 m/s + 0.0586 m/s^2 * t. Rearranging the equation to solve for t, we get t = (19.5 m/s) / (0.0586 m/s^2) ≈ 332.6 seconds.
Therefore, it takes approximately 332.6 seconds for the heavy freight train to increase its speed from 0 to 70.3 km/h.