Problem 28 - Master Resource Book in JEE Main Physics

Question

The engine of a train can impart a maximum acceleration of \(1 \mathrm{~ms}^{-2}\) and the brakes can give a maximum retardation of \(3 \mathrm{~ms}^{-2} .\) The least time during which a train can go from one place to the other place at a distance of \(1.2 \mathrm{~km}\) is nearly (a) \(108 \leq\) (b) \(191 \mathrm{~s}\) (c) \(56.6 \mathrm{~s}\) (d) time is fixed

Step-by-Step Solution

Answer

The least time taken is approximately \(46.19 \, \text{s}\), so \(56.6 \, \text{s}\) is appropriate.

Step 1: Understanding the problem

We are given that the train accelerates at a maximum rate of \(1 \, \text{ms}^{-2}\) and decelerates at a maximum rate of \(3 \, \text{ms}^{-2}\). The total distance to be covered is \(1.2\, \text{km} = 1200 \, \text{m}\). We need to calculate the shortest possible time for the train to travel this complete distance.

Step 2: Identify the phases of motion

The train will have two phases of motion: acceleration and deceleration. During acceleration, the velocity increases from 0 to some maximum velocity \(v\). During deceleration, the velocity decreases back to 0. The train covers half the distance in each phase.

Step 3: Calculate maximum velocity and time for each phase

Using the equation \(v^2 = u^2 + 2as\) where \(u = 0\), \(a = 1 \, \text{ms}^{-2}\), and \(s = 600 \, \text{m}\) (half of 1200 \(\text{m}\)), calculate \(v\): \[v^2 = 0 + 2 \times 1 \times 600 = 1200 \]\[v = \sqrt{1200} \approx 34.64 \, \text{ms}^{-1}\]Calculate the time \(t_a\) to reach \(v\) using \(v = at\):\[34.64 = 1 \times t_a \Rightarrow t_a \approx 34.64 \, \text{s}\]

Step 4: Calculate deceleration time

For braking, the deceleration rate is \(-3 \, \text{ms}^{-2}\). Using the same maximum velocity \(34.64 \, \text{ms}^{-1}\), calculate the time \(t_d\) to stop using \(v = u + at\), where \(v = 0\) and \(u = 34.64\):\[0 = 34.64 - 3t_d\]\[3t_d = 34.64 \Rightarrow t_d \approx 11.55 \, \text{s}\]

Step 5: Total time to cover the distance

Add the two time phases to find the total travel time: \[t = t_a + t_d = 34.64 + 11.55 = 46.19 \, \text{s}\]This is the time taken for acceleration and deceleration to cover the 1200 meter distance.

Key Concepts

Acceleration

Acceleration refers to how quickly an object's velocity changes over time. In the context of our train problem, the acceleration is the rate at which the train's speed increases as it begins its journey. The engine of the train can provide a maximum acceleration of \(1 \mathrm{\,ms}^{-2}\). This means that every second, the train's velocity increases by \(1 \mathrm{\,m/s}\). Acceleration can be calculated using the formula:

• \(a = \frac{\Delta v}{\Delta t}\)

This equation tells us that acceleration \(a\) is the change in velocity \(\Delta v\) over the change in time \(\Delta t\). In our problem, during acceleration, the train's velocity goes from \(0\) to approximately \(34.64 \, \text{ms}^{-1}\) over a certain time period, which we found to be about \(34.64 \, \text{s}\).
This increase in speed enables the train to cover the first half of the distance—\(600 \, \text{meters}\)—efficiently.

Deceleration

Deceleration is essentially the opposite of acceleration. It describes how quickly an object's velocity decreases over time. For the train, this occurs during the braking phase, where the brakes apply a maximum deceleration of \(-3 \mathrm{\,ms}^{-2}\). In simpler terms, the train's velocity decreases by \(3 \, \text{meters per second}\) every second as it comes to a stop. The formula for deceleration can be expressed similarly to acceleration:

• \(a = \frac{\Delta v}{\Delta t}\)

However, in this scenario, \(a\) will be negative, reflecting the slowdown.
In the exercise, the train decelerates from the velocity of approximately \(34.64 \, \text{ms}^{-1}\) back to \(0 \, \text{ms}^{-1}\). The time taken for this deceleration phase is about \(11.55 \, \text{seconds}\), allowing the train to cover the remaining \(600 \, \text{meters}\).
Understanding both acceleration and deceleration phases helps us see how motion is controlled to travel the total distance efficiently.

Motion Equations

Motion equations are the mathematical descriptions that relate the various factors involved in an object's movement—like displacement, velocity, acceleration, and time. A well-known set of these is the set of kinematic equations, and in our problem, we use these to calculate both acceleration and deceleration phases. The key equations used include:

• \(v^2 = u^2 + 2as\)
• \(v = u + at\)

In the calculation, \(v^2 = u^2 + 2as\) is used to find the maximum speed \(v\) the train reaches during the acceleration phase, while \(v = u + at\) is used to determine the time taken for these phases.
These equations assume a constant acceleration, which allows us to break down complex motions into manageable calculations. By understanding and applying these equations, one can solve a wide range of motion problems in physics.

Velocity

Velocity is a vector quantity, which means it has both magnitude and direction. It essentially indicates how fast an object is moving and in which direction. In the case of our train, velocity starts from zero as it begins, peaks when acceleration concludes, and returns to zero upon stopping. During acceleration, it increases to about \(34.64 \, \text{ms}^{-1}\) before the train starts to decelerate.
The velocity of the train is crucial since it determines how quickly it can traverse the given distance of \(1200 \text{m}\). By knowing the velocity in each phase, we can determine how long each phase will take. Velocity provides a clear picture of the train's speed at any point during the journey. Remember, since velocity includes direction, a change in direction (not in this scenario) would also mean a change in velocity, not just speed.