Question

A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform. The distance travelled by train for the above velocity is

• A. 3 km
• B. 5 km
• C. 8 km
• D. 12 km

Hint:

Using \(s = \frac{u + v}{2} \times t\) is often much faster than calculating acceleration first, especially when you don't need the acceleration value for anything else. Keep an eye on unit consistency (hours vs minutes).

Answer 1

The Correct Option is A

To find the distance travelled by the train with the given conditions, we will follow these steps:

1. First, convert the given velocity from km/h to m/s.

The velocity is given as 72 km/h. To convert km/h to m/s, use the conversion factor \(1 \, \text{km/h} = \frac{5}{18} \, \text{m/s}\).

Thus, the velocity \(v = 72 \times \frac{5}{18} \, \text{m/s} = 20 \, \text{m/s}\).

1. Determine the uniform acceleration of the train.

The train starts from rest, implying that the initial velocity \(u = 0\). The final velocity \(v = 20 \, \text{m/s}\) and the time taken to reach this velocity is 5 minutes, which is equivalent to \(300 \, \text{s}\).

Using the formula for acceleration \(a = \frac{v-u}{t}\), we get:

\(a = \frac{20 - 0}{300} = \frac{20}{300} \, \text{m/s}^2 = \frac{1}{15} \, \text{m/s}^2\).

1. Calculate the distance travelled using the equation of motion.

We use the equation \(s = ut + \frac{1}{2} a t^2\) to find the distance \(s\).

Substituting the values, \(u = 0\), \(a = \frac{1}{15} \, \text{m/s}^2\), and \(t = 300 \, \text{s}\):

\(s = 0 \cdot 300 + \frac{1}{2} \cdot \frac{1}{15} \cdot (300)^2\)

\(s = \frac{1}{2} \cdot \frac{1}{15} \cdot 90000\)

\(s = \frac{1}{2} \cdot 6000\)

\(s = 3000 \, \text{m} = 3 \, \text{km}\).

Therefore, the distance travelled by the train is 3 km. This confirms that the correct answer is 3 km.