91Ó°ÊÓ

In physics, power is the rate at which work is done or energy is transferred. For the locomotive, we use the power given, which is 1.5 MW, and convert it into watts for ease of calculation: \(1.5 \text{ MW} = 1.5 \times 10^6 \text{ W}\).
Using the formula \(P = F \times v\), where \(P\) is power, \(F\) is force, and \(v\) is velocity, we can find the force when the velocity is known. This relationship helps us solve various parts of the problem, including calculating the mass where power is directly linked to mass through acceleration and velocity.

Acceleration is the rate of change of velocity with time. For our locomotive, we have the initial speed \(v_i = 10 \text{ m/s}\) and the final speed \(v_f = 25 \text{ m/s}\).
The time interval for this change is given as 6 minutes, which we need to convert into seconds: \(6 \text{ min} = 6 \times 60 \text{ s} = 360 \text{ s}\).
Using the formula for acceleration, \( a = \frac{v_f - v_i}{\Delta t}\), we plug in the values: \(a = \frac{25 \text{ m/s} - 10 \text{ m/s}}{360 \text{ s}} = 0.04167 \text{ m/s}^2\).
This constant acceleration helps us to solve other parts of the problem, such as the change in speed over time and the distance traveled.

Kinematics involves the study of motion without considering its causes. We use it here to determine how the train's speed changes over time and how far it travels during the acceleration phase.
Using the equation for speed as a function of time \(v(t) = v_i + a \times t\), we find the speed at any given time \((t)\). Given our initial velocity (\ref{v_i = 10 \text{ m/s}}), acceleration (\ref{a = 0.04167 \text{ m/s}^2}), and the general form \(v(t) = 10 \text{ m/s} + 0.04167 \text{ m/s}^2 \times t\). So, for instance, at \(t = 180 \text{ s}\), the speed is \(v(180) = 10 + 0.04167 \times 180 ≈ 17.5 \text{ m/s}\). Similarly, the distance traveled can be calculated using \(s = v_i \times t + \frac{1}{2} \times a \times t^2\). Plugging in our values for a total time of 360 seconds, we get \(s = 10 \times 360 + \frac{1}{2} \times 0.04167 \times 360^2 ≈ 6300 \text{ m}\).

To find the mass of the train, we use the relationship between power, force, and velocity. From the power formula \(P = F \times v\), we can find force (\ref{F = m \times a}). Rearrange to solve for mass \(m = \frac{P}{a \times v}\).
We calculate the average velocity first \(\overline{v} = \frac{v_i + v_f}{2} = \frac{10 + 25}{2} = 17.5 \text{ m/s}\).
Now, inserting the given power and calculated values for acceleration and average velocity, we get: \(m = \frac{1.5 \times 10^6 \text{ W}}{0.04167 \text{ m/s}^2 \times 17.5 \text{ m/s}} ≈ 2.059 \times 10^6 \text{ kg}\). Thus, the mass of the train is approximately 2.059 million kilograms.

The distance traveled by the train during its acceleration phase can be found using the formula for motion under constant acceleration: \(s = v_i \times t + \frac{1}{2} \times a \times t^2\). Here, initial velocity \(v_i = 10 \text{ m/s}\), acceleration \(a = 0.04167 \text{ m/s}^2\), and time \(t = 360 \text{ s}\): \(s = 10 \text{ m/s} \times 360 \text{ s} + \frac{1}{2} \times 0.04167 \text{ m/s}^2 \times (360 \text{ s})^2\).
Calculating this gives: \(s = 3600 \text{ m} + 2700 \text{ m} = 6300 \text{ m}\).
So, the train travels a total distance of 6300 meters during the 6 minutes of acceleration.