Problem 121 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration \(\mathrm{f}\) and in the second part with constant retardation \(\mathrm{r}\). The value of \(\mathrm{t}\) is given by \([2003]\) (a) \(\sqrt{2 s\left(\frac{1}{f}+\frac{1}{r}\right)}\) (b) \(2 s\left(\frac{1}{f}+\frac{1}{r}\right)\) (c) \(\frac{2 s}{\frac{1}{f}+\frac{1}{r}}\) (d) \(\sqrt{2 s(f+r)}\)

Step-by-Step Solution

Answer

The correct answer is option (a): \( \sqrt{2s(\frac{1}{f} + \frac{1}{r})} \).

Step 1: Identify the components of motion

The motion of the body is divided into two phases. During the first phase, it undergoes constant acceleration \( f \), and during the second phase, it experiences constant retardation (deceleration) \( r \). The body starts from rest and comes to rest at the end of the journey.

Step 2: Apply concepts of uniform acceleration

For the first part of the journey where the body accelerates, use the equation: \( v = ft \) (since initial velocity \( u = 0 \)). The velocity \( v \) at the end of the first phase is \( ft \).

Step 3: Calculate distance covered during acceleration

The distance \( s_1 \) covered during the acceleration phase can be calculated using the equation: \( s_1 = \frac{1}{2} f t_1^2 \) where \( t_1 \) is the time for acceleration.

Step 4: Evaluate distance during retardation

In the second part, the body is moving with deceleration \( r \) and it starts from velocity \( ft \) acquired from the first part and comes to rest. The time \( t_2 \) for this phase can be calculated as \( t_2 = \frac{ft}{r} \).

Step 5: Calculate distance during retardation

The distance \( s_2 \) covered during retardation is calculated as: \( s_2 = \frac{v^2}{2r} = \frac{(ft)^2}{2r} \).

Step 6: Total distance equation

The total distance \( s \) covered is the sum of the distances covered in each part: \( s = s_1 + s_2 \), so \( s = \frac{1}{2} ft_1^2 + \frac{f^2t_2^2}{2r} \).

Step 7: Solve for total time and match given options

We know \( t = t_1 + t_2 \). Use \( t_1 = \frac{v}{f} \) and \( t_2 = \frac{v}{r} \). Calculate \( t = \frac{v}{f} + \frac{v}{r} = v(\frac{1}{f} + \frac{1}{r}) \). Match this with the given options to find that \( t = \sqrt{2s(\frac{1}{f} + \frac{1}{r})} \).

Step 8: Identify the correct answer option

Ultimately, the correct relation for time \( t \) is: \( \sqrt{2s(\frac{1}{f} + \frac{1}{r})} \). This matches with option (a).

Key Concepts

Constant Acceleration

When an object experiences constant acceleration, its velocity increases steadily over time. This type of motion is often described by the formula:

• \( v = u + ft \)

where \( v \) is the final velocity, \( u \) is the initial velocity, \( f \) is the acceleration, and \( t \) is the time.
This equation tells us how quickly an object's velocity changes when it accelerates constantly. In many physics problems, like the one here, you might start observing an object from rest, meaning the initial velocity \( u \) is zero. This simplifies the equation to \( v = ft \).

Constant acceleration can occur in everyday scenarios, like a car speeding up on a straight road. It's important to understand that constant acceleration means the rate of change of velocity is uniform.
This simplicity makes it easier to predict the object's future position and velocity, using additional formulas such as:

• \( s = ut + \frac{1}{2}ft^2 \)

Here, \( s \) denotes the distance covered during the acceleration.

Uniform Acceleration

Uniform acceleration is another term used interchangeably with constant acceleration. When acceleration is uniform, besides a steady rate of velocity change, the motion follows predictable patterns that can be described using basic equations of motion. This makes problem-solving straightforward.

In situations involving uniform acceleration, both the magnitude and direction of acceleration remain the same throughout the motion. Consider situations in physics problems:

• Throwing an object upwards where it faces uniform gravitational pull
• Sliding an object on a frictional surface with constant resistance

In any case described as having uniform acceleration, you can rely on solving it with kinematic equations like the ones mentioned in the constant acceleration section.

Also, understanding the nature of uniform acceleration helps in grasping the relationship between motion parameters like distance, velocity, and time, which is pivotal in physics.

Retardation

Retardation refers to a uniform negative acceleration, where an object slows down over time. It is also known as deceleration. In many exercises, this is crucial for understanding phases of motion where an object comes to a stop.
Retardation can be calculated similar to acceleration, but with its value being negative. For example, if during retardation the velocity is reduced uniformly from \( v \) to zero, the event uses this concept:

• Formula: \( v = u - rt \)

Here, \( u \) is the initial velocity, \( r \) is the retardation, and \( t \) is the time taken to slow down.

In physics exercises involving both acceleration and retardation phases, like the original one, you manage the calculations for each part separately. Calculating the correct time period where retardation occurs and combining it with acceleration computations often solves such problems.
Practically, understanding retardation helps comprehend how vehicles slow down, aiding not just in academics but in real-world driving scenarios too.