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?
m = 14000 MT * 1000 kg/MT = 1.4*10^7 kg
MT = Metric Tons. m = mass.
F = ma,
a = F/m = 6.45*10^5 / 1.4*10^7 = 0.0457m/s^2.
V = 69.8 km/h * 1000 m/km * 1/3600 h/s =19.39 m/s.
a = (Vf - Vo) / t = 0.0457,
(19.39 - 0) / t = 0.0457,
19.39 / t = 0.0457,
Cross multiply:
0.0457t = 19.39,
t = 19.39 / 0.0457 = 424.3 s = 7.1 min.
Well, well, well, looks like we have a heavy train trying to get up to speed! Let's crunch those numbers and find out how long it'll take.
First things first, we need to convert that speed from kilometers per hour to meters per second to keep our physics in check. So, 69.8 km/h would be approximately 19.4 m/s. Don't worry, I'm not derailing this joke!
Now, let's get back on track and apply Newton's Second Law: force equals mass times acceleration. In this case, our acceleration is the change in speed per second.
So, rearranging the formula, we have acceleration equals force divided by mass. In this case, acceleration comes out to be approximately 46.07 m/s^2. Now, we just need to find out how long it takes for the train to reach 19.4 m/s with this acceleration.
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is acceleration, and t is time, we can plug in the numbers.
0 m/s (initial velocity) + 46.07 m/s^2 (acceleration) x t (time) = 19.4 m/s (final velocity)
Doing some calculations, we find that it takes approximately 0.42 seconds for this heavy freight train to reach a speed of 19.4 m/s.
And there you have it! In just about 0.42 seconds, that train will have chugged its way up to 69.8 km/h. Now, that's what I call locomotive speed!
To solve this problem, we will use Newton's second law of motion: Force = mass × acceleration. We need to find the acceleration of the train first.
Given:
Mass of the freight train = 14000 metric tons = 14000 × 1000 kg = 14,000,000 kg
Force applied by the locomotive (F) = 645,000 N
Initial velocity (u) = 0 km/h
Final velocity (v) = 69.8 km/h
First, we need to convert the velocities from km/h to m/s:
Final velocity (v) = 69.8 km/h × (1000 m/1 km) × (1 h/3600 s) = 19.4 m/s
We can calculate the acceleration (a) using the formula:
Force (F) = mass (m) × acceleration (a)
Acceleration (a) = Force (F) / mass (m)
a = 645,000 N / 14,000,000 kg
a ≈ 0.0461 m/s²
Next, we will use the equation of motion to calculate the time it takes for the train to accelerate from rest (initial velocity = 0 m/s) to the final velocity (v = 19.4 m/s).
v = u + at
19.4 m/s = 0 m/s + 0.0461 m/s² × t
t = 19.4 m/s / 0.0461 m/s²
t ≈ 421.08 seconds
Therefore, it would take approximately 421.08 seconds to increase the train's speed from 0 to 69.8 km/h.
To find the time it takes for the freight train to increase its speed from 0 to 69.8 km/h, we can use Newton's second law of motion.
Newton's second law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). Rearranging this equation, we can solve for acceleration as acceleration = F/mass.
First, let's convert the mass of the freight train from metric tons to kilograms. Since 1 metric ton is equal to 1000 kilograms, the mass of the train is 14,000 metric tons * 1000 kg/ton = 14,000,000 kg.
Next, we need to convert the speed from km/h to m/s, as the SI unit for acceleration is meters per second squared (m/s^2). We can convert km/h to m/s by multiplying the given speed by 1000/3600, since there are 1000 meters in a kilometer and 3600 seconds in an hour. So, the speed of the train is 69.8 km/h * 1000 m/km / 3600 s/h = 19.4 m/s.
Now, we can substitute the force and mass values into the equation acceleration = F/mass:
acceleration = 645,000 N / 14,000,000 kg
acceleration ≈ 0.0461 m/s^2
Since we know the initial speed is 0 m/s, the final speed is 19.4 m/s, and the acceleration is 0.0461 m/s^2, we can use the equation of motion vf = vi + at to find the time (t) it takes for the train to increase its speed. Rearranging the equation, we get t = (vf - vi) / a.
Plugging in the values:
t = (19.4 m/s - 0 m/s) / 0.0461 m/s^2
t ≈ 421.67 seconds
Therefore, it takes approximately 421.67 seconds for the heavy freight train to increase its speed from 0 to 69.8 km/h.