A freight train has a mass of 1.7 × 10^7 kg.
If the locomotive can exert a constant pull
of 5.2 × 10^5 N, how long would it take to
increase the speed of the train from rest to
74.6 km/h? Disregard friction.
Answer in units of s
To find the time it takes for the train to accelerate from rest to a certain speed, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = ma).
In this case, the force acting on the train is the pull exerted by the locomotive, which is given as 5.2 × 10^5 N. The mass of the train is 1.7 × 10^7 kg. The acceleration of the train can be calculated using the formula:
acceleration = force / mass
Substituting the given values:
acceleration = (5.2 × 10^5 N) / (1.7 × 10^7 kg)
Next, we need to find the distance traveled by the train during the acceleration period. We can use the equation:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Since the train starts from rest, the initial velocity is 0. We are given the final velocity, which is 74.6 km/h. However, we need to convert it to m/s before using it in the equation.
Conversion:
1 km/h = 1000 m/3600 s = 5/18 m/s
So, the final velocity is:
final velocity = 74.6 km/h * (5/18 m/s) = 20.7 m/s
Plugging in the values, we have:
distance = (0 * time) + (0.5 * acceleration * time^2)
distance = 0.5 * acceleration * time^2
Now we can use the equation for distance and solve for time:
distance = 0.5 * acceleration * time^2
time^2 = (2 * distance) / acceleration
time = sqrt((2 * distance) / acceleration)
Substituting the calculated values:
time = sqrt((2 * distance) / acceleration)
time = sqrt((2 * 0) / acceleration)
time = sqrt(0)
time = 0
Therefore, the time it takes for the train to accelerate from rest to 74.6 km/h is 0 seconds.