A freight train has a mass of 3.1 × 107 kg.

If the locomotive can exert a constant pull
of 4.9 × 105 N, how long would it take to
increase the speed of the train from rest to
71.9 km/h? Disregard friction.

a=F/m=4.9*10^5 / 3.1*10^7=0.0158 m/s^2.

V = 71.9km/h = 71900m/3600s = 20 m/s.

t = (V-Vo)/a = (20-0)/0.0158 = 1265.8 s.
= 21 min.

To find the time it takes to accelerate the train from rest to a certain speed, we can solve this problem using Newton's second law of motion.

Step 1: Convert the speed of the train from km/h to m/s.
To convert km/h to m/s, we need to divide the speed by 3.6 since there are 3.6 meters in one second.
71.9 km/h divided by 3.6 = 19.972 m/s (rounded to three decimal places).

Step 2: Calculate the net force acting on the train.
The net force acting on the train is the force produced by the locomotive minus the force due to inertia.
The force produced by the locomotive is given as 4.9 × 10^5 N.

Step 3: Calculate the force due to inertia.
The force due to inertia is equal to the mass of the train multiplied by its acceleration.
Since the train is initially at rest, the initial velocity is 0 m/s.
Using the equation, F = m * a, we can solve for acceleration (a):
4.9 × 10^5 N = (3.1 × 10^7 kg) * a.
Solving for a: a = (4.9 × 10^5 N) / (3.1 × 10^7 kg) = 0.0158 m/s^2 (rounded to four decimal places).

Step 4: Calculate the time taken to accelerate.
To calculate the time taken to accelerate, we can use the equation:
v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
In this case, u = 0 m/s, v = 19.972 m/s, and a = 0.0158 m/s^2.

Rearranging the equation to solve for time, we have:
t = (v - u) / a

Plugging in the values:
t = (19.972 m/s - 0 m/s) / 0.0158 m/s^2
t = 1263.924 s (rounded to three decimal places).

Therefore, it would take approximately 1263.924 seconds (or 21 minutes and 3.924 seconds) to increase the speed of the train from rest to 71.9 km/h.