A heavy freight train has a mass of 14000 metric tons. If the locomotive can pull with a force of 645000 N, how long does it take to increase it's speed from 0 to 69.8 km/h?
To work with consistent units, you need to convert 69.8 km/h to m/s, namely
69.8 km/h * 1000 m/km / 3600 s/h
= 69.8*1000/3600 km*m*h/(h*km*s)
= 19.39 m/s
After that, solve for a in F=ma, where F and m are known.
The time required in seconds can be obtained by dividing 19.39 m/s by the acceleration in m/s².
To find out how long it takes for the train to increase its speed, we can use the concept of Newton's second law of motion, which states that F = ma, where F is the force, m is the mass, and a is the acceleration.
First, let's convert the train's mass from metric tons to kilograms. Since 1 metric ton is equal to 1000 kilograms, the train's mass is 14,000 * 1000 = 14,000,000 kilograms.
Next, let's convert the speed from km/h to m/s, as the unit of force is in Newtons (N) and the unit of acceleration is in m/s^2. To convert, we divide 69.8 km/h by 3.6 to get the speed in m/s: 69.8 / 3.6 = 19.4 m/s.
Now, the initial speed is 0 m/s, and we want to increase it to 19.4 m/s. So the change in speed (∆v) is 19.4 m/s - 0 m/s = 19.4 m/s.
Since acceleration (a) is the change in velocity divided by the time taken (∆t), we can rearrange the formula a = ∆v / ∆t to solve for time (∆t).
In this case, a = F / m, where F is the force exerted by the locomotive (645,000 N) and m is the mass of the train (14,000,000 kg).
Substituting the values into the formula:
645,000 N / 14,000,000 kg = 19.4 m/s / ∆t
Let's calculate ∆t:
∆t = (14,000,000 kg * 19.4 m/s) / 645,000 N
Simplifying:
∆t = 419,600,000 kg⋅m/s / 645,000 N
∆t ≈ 650.39 seconds
Therefore, it takes approximately 650.39 seconds (or about 10.84 minutes) for the heavy freight train to increase its speed from 0 to 69.8 km/h.