Problem 68 - Master Resource Book in JEE Main Physics

Question

An express train is moving with a velocity \(v_{1}\) its driver finds another train is moving on the same track in the same direction with velocity \(v_{2}\). To avoid collision driver applies a retardation \(a\) on the train. The minimum time of avoiding collision will be (a) \(t=\frac{v_{1}-v_{2}}{a}\) (b) \(t=\frac{v_{1}^{2}-v_{2}^{2}}{2}\) (c) None (d) Both (a) and (b)

Step-by-Step Solution

Answer

(a) \( t = \frac{v_1 - v_2}{a} \)

Step 1: Identify the Problem

We need to determine the minimum time required for one train, initially traveling with velocity \( v_1 \), to decelerate (or apply negative acceleration, \( a \)) and avoid colliding with a second train moving at velocity \( v_2 \) in the same direction.

Step 2: Formulate the Relative Speed

Since both trains are moving in the same direction, the relative speed \( v_r \) of the express train with respect to the second train is given by the difference of their speeds: \[ v_r = v_1 - v_2. \] This `relative speed` is the initial speed for solving the collision problem.

Step 3: Use the Equation of Motion for Retardation

To avoid the collision, the express train must reduce its speed relative to the other train to zero. The equation for motion with constant negative acceleration (retardation) is: \[ v = u + at, \] where \( v \) is the final velocity (zero for avoiding collision), \( u \) is the initial velocity (relative speed \( v_r \)), \( a \) is the retardation (negative acceleration), and \( t \) is the time.

Step 4: Solve for Time

Substitute the values into the motion equation: set \( v = 0 \), \( u = v_1 - v_2 \), and \( a \) as \( -a \) for deceleration. The equation becomes: \[ 0 = (v_1 - v_2) - at. \] Solving for \( t \), we get: \[ t = \frac{v_1 - v_2}{a}. \] This matches option (a).

Step 5: Verify and Choose the Correct Option

Given our derived result \( t = \frac{v_1 - v_2}{a} \), option (a) is correct. Option (b) \( t = \frac{v_1^2 - v_2^2}{2} \) doesn't match our calculations and thus is incorrect. Hence, the correct answer is only (a).

Key Concepts

Relative Speed

Relative speed is an essential concept when discussing collision avoidance between two moving objects. It refers to the speed of one object as observed from another moving object. In this exercise, two trains are moving on the same track in the same direction. The express train moves at a speed of \( v_1 \) while the other train travels at a speed of \( v_2 \). To find out how quickly one train approaches the other, we calculate the relative speed: \( v_r = v_1 - v_2 \).

This calculation helps to determine how fast the express train is closing the gap between itself and the other train.

• If \( v_1 \) is significantly larger than \( v_2 \), the express train will be approaching fast and hence, needs immediate action to avoid a collision.
• If \( v_1 \) is only slightly greater than \( v_2 \), the express train is closing the distance more slowly, allowing more time for collision avoidance.

Understanding relative speed is crucial because it sets the stage for determining the necessary actions to prevent a collision.

Equation of Motion

The equation of motion is a powerful tool used to predict the future position or speed of an object. In our problem, we deal with a specific application of this equation due to the need for collision avoidance. The express train needs to reduce its relative speed from \( v_r \) to zero, meaning it must match the speed of the other train.

The equation of motion used here is: \ \( v = u + at \),
where \( v \) is the final velocity, \( u \) is the initial velocity (in our case, the relative speed \( v_r \)), \( a \) is the acceleration (here, a negative value called retardation since the train decelerates), and \( t \) is the time.

• By setting \( v \) to zero (as the trains should have matching speeds), we rearrange the equation to solve for time \( t \).
• This lets us find out how long it will take for the express train to slow down enough to match speeds with the other train, thereby avoiding a collision.

Using this equation efficiently dictates the actions needed in real life to ensure safety on the tracks.

Retardation

Retardation is the term used to describe negative acceleration. It denotes a decrease in speed over time. In the context of our exercise, the express train's driver applies a retardation \( a \) to reduce speed and avoid collision.

When we encounter retardation:

• The train needs to slow down quickly enough to prevent a crash while staying within safe limits of force application.
• This parameter is crucial in calculating how quickly the train can decelerate to match the slower train's speed.

Understanding retardation is vital because it affects the safety and comfort of passengers. It also determines the braking system's efficiency and effectiveness, linking directly back to our main concern—avoiding collisions through calculated slowing.

Constant Acceleration

Constant acceleration means the acceleration value does not change over time. Though 'acceleration' typically implies speeding up, when the value is negative, as in our train problem, it represents constant deceleration or retardation.

The benefit of constant acceleration (or deceleration) in solving the exercise is that it simplifies calculations:

• This consistency allows us to apply the equation of motion effectively without having to account for varying changes in acceleration.
• In practical scenarios, it helps train systems design braking mechanisms that provide predictable stopping patterns, enhancing safety.

Constant acceleration ensures that the mathematical relationships, like \( v = u + at \), hold true, making predictions more straightforward and reliable.