Problem 50 - Master Resource Book in JEE Main Physics

Question

A train accelerated uniformly from rest attains a maximum speed of \(40 \mathrm{~ms}^{-1}\) in \(20 \mathrm{~s}\). It travels at this speed for \(20 \mathrm{~s}\) and is brought to rest with uniform retardation in \(40 \mathrm{~s}\). The average velocity during this period is (a) \((80 / 3) \mathrm{ms}^{-1}\) (b) \(30 \mathrm{~ms}^{-1}\) (c) \(25 \mathrm{~ms}^{-1}\) (d) \(40 \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

The average velocity is 25 ms^{-1}, so the answer is (c).

Step 1: Calculate the distance during acceleration

The train accelerates uniformly from rest (initial velocity, \(u = 0\)) to a maximum speed \(v = 40 \mathrm{~ms}^{-1}\) over \(t = 20 \mathrm{~s}\). The formula for distance \(s\) under uniform acceleration is \(s = ut + \frac{1}{2}at^2\). Here, since \(u = 0\), it simplifies to \(s = \frac{1}{2}at^2\). We first find the acceleration using \(v = u + at,\) which gives \(40 = 0 + a \times 20\), so \(a = 2 \mathrm{~ms}^{-2}\). Now, substitute \(a\) and \(t\) into the distance formula: \(s_1 = \frac{1}{2} \times 2 \times 20^2 = 400 \mathrm{~m}\).

Step 2: Calculate the distance at constant speed

The train then travels at a constant speed \(v = 40 \mathrm{~ms}^{-1}\) for \(20 \mathrm{~s}\). The distance \(s\) travelled at constant speed is \(s = vt\). Substitute the values: \(s_2 = 40 \times 20 = 800 \mathrm{~m}\).

Step 3: Calculate the distance during deceleration

The train slows down from \(40 \mathrm{~ms}^{-1}\) to rest over \(40 \mathrm{~s}\) with uniform retardation. Using the same formula for distance \(s = ut + \frac{1}{2}at^2\), where \(u = 40 \mathrm{~ms}^{-1}\) and \(v = 0\), we calculate the retardation first: Using \(v = u + at\), \(0 = 40 + a \times 40\), so \(a = -1 \mathrm{~ms}^{-2}\). Substitute \(u\), \(a\), and \(t\) in the distance formula: \(s_3 = 40 \times 40 + \frac{1}{2} \times (-1) \times 40^2 = 800 \mathrm{~m}\).

Step 4: Calculate total distance and time

Sum up all distances: \(s = s_1 + s_2 + s_3 = 400 + 800 + 800 = 2000 \mathrm{~m}\). Sum the times for each segment: Total time, \(t = 20 + 20 + 40 = 80 \mathrm{~s}\).

Step 5: Find the average velocity

The average velocity is given by the total distance divided by the total time. \(v_{\text{avg}} = \frac{s}{t} = \frac{2000}{80} = 25 \mathrm{~ms}^{-1}\). Compare this result to the options provided to find the correct answer.

Key Concepts

Uniform Acceleration

Uniform acceleration occurs when an object's velocity changes at a constant rate over time. It means that the acceleration does not vary as time progresses. This concept is crucial in the study of motion in kinematics.

For instance, consider a train that accelerates uniformly from rest. The initial velocity (\(u\)) is 0 \(\text{ms}^{-1}\), meaning it starts from a standstill. The train reaches a final velocity (\(v\)) after a certain time period (\(t\)). In the example, the train reaches a velocity of \(40 \text{ms}^{-1}\) after 20 seconds.

We use the equation \(v = u + at\) to calculate the acceleration (\(a\)) during this period. Substituting the known values, we find: \(40 = 0 + a \times 20\), giving an acceleration of \(2 \text{ms}^{-2}\).

Understanding uniform acceleration is essential because it simplifies calculations for velocity and distance, helping us predict how an object will move over time.

Average Velocity

Average velocity is a way to describe the motion of an object over a period of time, taking into account its total displacement and the total time taken. This is a central concept in analyzing how objects move.

The formula to determine average velocity (\(v_{\text{avg}}\)) is \(v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}\). In this scenario, the train covers different distances during its journey, including when it accelerates, moves at constant speed, and decelerates.

The train's average velocity can be found, even if its speed was not constant throughout. By summing all the distances (total distance = 2000 meters) and dividing by the total time (80 seconds), we find that the average velocity \(v_{\text{avg}} = \frac{2000}{80} = 25 \text{ms}^{-1}\).

This concept helps us make sense of varied motions and provides a single value to represent an object's entire journey.

Distance Calculation

Distance calculation is a fundamental aspect in kinematics, enabling us to determine how far an object has traveled throughout different phases of its motion.

For uniform acceleration, the distance (\(s\)) covered can be found using the formula \(s = ut + \frac{1}{2}at^2\), where \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. In the provided example, the train's distance during acceleration is calculated by substituting known values into this formula, resulting in a distance of 400 meters.

When an object moves at a constant speed, the distance is simply \(s = vt\), where \(v\) is the constant velocity and \(t\) is the time. For the train traveling at a constant speed of \(40 \text{ms}^{-1}\), over 20 seconds, the distance is 800 meters.

During deceleration, we again use the formula for uniform acceleration/deceleration to find the distance. The constant application of these formulas helps us to derive consistent results in various motion segments.