A freight train has a mass of 1.8 × 10
7
kg.
If the locomotive can exert a constant pull
of 6.2×10
5
N, how long does it take to increase
the speed of the train from rest to 70 km/h?
Answer in units of min.
F = m a
6.2 * 10^5 = 1.8 * 10^7 a
a = (6.2/1.8)10^-2 m/s^2
v = a t
now what is v
70,000 m/3600 s = 19.4 m/s
so
t = (19.4 * 1.8 /6.2 )* 10^2 seconds
To find the time it takes for the train to accelerate from rest to 70 km/h, we can use the equation:
F = ma
where F is the force exerted by the locomotive, m is the mass of the train, and a is the acceleration.
First, let's convert the speed from km/h to m/s:
70 km/h = 70,000 m/3600 s ≈ 19.44 m/s
Since the train starts from rest, its initial velocity (u) is 0 m/s. The final velocity (v) is 19.44 m/s, and the displacement (s) is not given.
We can use the formula:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the train starts from rest, the initial velocity is 0 m/s, so the formula becomes:
v = at
Rearranging the formula to solve for acceleration:
a = v / t
Substituting the given values:
6.2 × 10^5 N = (1.8 × 10^7 kg) * (19.44 m/s) / t
Solving for t:
t = (1.8 × 10^7 kg * 19.44 m/s) / (6.2 × 10^5 N)
t ≈ 571 seconds
However, the answer is required in units of minutes. To convert seconds to minutes:
t = 571 seconds * (1 minute / 60 seconds)
t ≈ 9.5 minutes
Therefore, it would take approximately 9.5 minutes to increase the speed of the train from rest to 70 km/h.
To find the time it takes to increase the speed of the train from rest to 70 km/h, we can use the equation of motion:
v = u + at
Where:
v = final velocity (70 km/h)
u = initial velocity (0 km/h, as the train is at rest)
a = acceleration (to be determined)
t = time (unknown)
First, let's convert the given velocities to m/s, as it is the standard SI unit for velocity. There are 1000 meters in 1 kilometer and 3600 seconds in 1 hour, so:
70 km/h = (70 * 1000) / 3600 = 19.44 m/s
Now, we need to find the acceleration, which can be calculated using Newton's second law of motion:
F = ma
Where:
F = force applied by the locomotive (6.2 × 10^5 N)
m = mass of the train (1.8 × 10^7 kg)
a = acceleration (to be determined)
Rearranging the formula to solve for acceleration:
a = F / m
a = (6.2 × 10^5 N) / (1.8 × 10^7 kg) = 0.0344 m/s^2
Now that we have the acceleration, we can substitute the values into the equation of motion and solve for time:
19.44 m/s = 0 + 0.0344 m/s^2 * t
Simplifying the equation:
19.44 = 0.0344t
Dividing both sides by 0.0344:
t = 19.44 / 0.0344 ≈ 565.12 seconds
Lastly, we need to convert the time from seconds to minutes by dividing by 60:
t = 565.12 / 60 ≈ 9.42 minutes
Therefore, it takes approximately 9.42 minutes to increase the speed of the train from rest to 70 km/h.