Understanding Train Motion and Distance Calculation

This problem asks us to find the distance a train travels while accelerating uniformly from rest to a specific velocity over a given time. To solve this, we need to use the equations of motion for uniformly accelerated bodies.

Identifying Given Information

• Initial velocity of the train ($u$) = starting from rest = $0 \text{ m/s}$
• Final velocity of the train ($v$) = $90 \text{ km/h}$
• Time taken ($t$) = $5 \text{ minutes}$
• Acceleration is uniform.

We need to find the distance travelled ($s$) during this time.

Units Conversion for Consistent Calculations

Before using kinematic equations, we must ensure all quantities are in consistent units. Let's convert the final velocity from kilometers per hour (km/h) to meters per second (m/s) and the time from minutes to seconds (s), which are the standard SI units.

• Converting Final Velocity:

  $$v = 90 \text{ km/h}$$

  Since $1 \text{ km} = 1000 \text{ m}$ and $1 \text{ hour} = 3600 \text{ seconds}$, we have:

  $$v = 90 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 90 \times \frac{10}{36} \text{ m/s} = 90 \times \frac{5}{18} \text{ m/s} = 5 \times 5 \text{ m/s} = 25 \text{ m/s}$$

• Converting Time:

  $$t = 5 \text{ minutes}$$

  Since $1 \text{ minute} = 60 \text{ seconds}$, we have:

  $$t = 5 \times 60 \text{ s} = 300 \text{ s}$$

So, the train starts at $u = 0 \text{ m/s}$, reaches $v = 25 \text{ m/s}$ in $t = 300 \text{ s}$ with uniform acceleration.

Calculating the Uniform Acceleration

We can use the first equation of motion, which relates initial velocity, final velocity, acceleration, and time:

$$v = u + at$$

Plugging in the values we have:

$$25 \text{ m/s} = 0 \text{ m/s} + a \times 300 \text{ s}$$

$$25 = 300a$$

Solving for acceleration ($a$):

$$a = \frac{25}{300} \text{ m/s}^2 = \frac{1}{12} \text{ m/s}^2$$

The uniform acceleration of the train is $\frac{1}{12} \text{ m/s}^2$.

Calculating the Distance Travelled

Now we can find the distance travelled ($s$) using the second equation of motion:

$$s = ut + \frac{1}{2}at^2$$

Plugging in the values for $u$, $a$, and $t$:

$$s = (0 \text{ m/s}) \times (300 \text{ s}) + \frac{1}{2} \times \left(\frac{1}{12} \text{ m/s}^2\right) \times (300 \text{ s})^2$$

$$s = 0 + \frac{1}{2} \times \frac{1}{12} \times (300)^2 \text{ m}$$

$$s = \frac{1}{24} \times (90000) \text{ m}$$

$$s = \frac{90000}{24} \text{ m}$$

Let's simplify the fraction:

$$s = \frac{45000}{12} \text{ m} = \frac{22500}{6} \text{ m} = \frac{11250}{3} \text{ m} = 3750 \text{ m}$$

Converting Distance back to Kilometers

The distance calculated is $3750$ meters. The options are given in kilometers. To convert meters to kilometers, we divide by 1000 (since $1 \text{ km} = 1000 \text{ m}$).

$$s = \frac{3750}{1000} \text{ km} = 3.75 \text{ km}$$

The distance travelled by the train is $3.75 \text{ km}$.

Summary of Calculation Steps:

1. Convert given velocity and time to consistent units (m/s and s).
2. Calculate uniform acceleration using $v = u + at$.
3. Calculate distance travelled using $s = ut + \frac{1}{2}at^2$.
4. Convert the final distance to the required unit (km).

Checking the Options

The calculated distance is $3.75 \text{ km}$. Let's compare this with the provided options:

• 1.5 km
• 3.25 km
• 2.25 km
• 3.75 km

The calculated distance matches the fourth option.

Revision Table: Key Kinematic Equations

Additional Information on Train Motion and Kinematics

Uniform Acceleration: This means the velocity of the train changes by the same amount in every equal interval of time. In this case, the acceleration $a$ is constant.

Starting from Rest: This is a specific initial condition where the initial velocity ($u$) is zero. This simplifies the kinematic equations, as terms containing $u$ often become zero.

Importance of Consistent Units: Using consistent units throughout the calculation is crucial. Mixing units (like km/h with minutes) without proper conversion will lead to incorrect results. The SI system (meters, kilograms, seconds) is generally preferred in physics problems.

Kinematic Equations: These equations are fundamental tools for analyzing motion under constant acceleration. Choosing the correct equation depends on the variables given and the variable you need to find.