Problem 6 - Master Resource Book in JEE Main Physics

Question

A car moving along a straight highway with speed of \(126 \mathrm{~km} / \mathrm{h}\) is brought to a stop within a distance of \(200 \mathrm{~m}\). What is the retardation of the car (assumed uniform) and how long does it take for the car to stop? [NCERT] (a \(3.06 \mathrm{~m} / \mathrm{s}^{2}\) and \(11.4 \mathrm{~s}\) (b) \(206 \mathrm{~m} / \mathrm{s}^{2}\) and \(11.4 \mathrm{~s}\) (c) \(3.06 \mathrm{~m} / \mathrm{s}^{2}\) and \(10.4 \mathrm{~s}\) (d) \(3.06 \mathrm{~m} / \mathrm{s}^{2}\) and \(4.1 \mathrm{~s}\)

Step-by-Step Solution

Answer

(a) \(3.06 \, \text{m/s}^2\) retardation and \(11.4 \, \text{s}\) time.

Step 1: Identify the Known Values

The initial speed \( u \) of the car is \( 126 \, \text{km/h} \). Convert this to meters per second by using the conversion factor (\( 1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s} \)). So, \( u = \frac{126}{3.6} \, \text{m/s} \). The final speed \( v \) is \( 0 \, \text{m/s} \) since the car stops. The distance \( s \) is \( 200 \, \text{m} \).

Step 2: Convert Initial Speed to Meters per Second

To convert \( 126 \text{ km/h} \) to \( \text{m/s} \), use the formula: \( u = \frac{126}{3.6} \). Calculate \( u \): \( \frac{126}{3.6} = 35 \, \text{m/s} \).

Step 3: Apply the Equation for Retardation

Use the formula \( v^2 = u^2 + 2as \) and rearrange it to solve for \( a \): \( a = \frac{v^2 - u^2}{2s} \). Substitute the known values: \( a = \frac{0 - (35)^2}{2 \times 200} = \frac{-1225}{400} \).

Step 4: Calculate the Retardation

Perform the division for the retardation: \( a = \frac{-1225}{400} = -3.0625 \, \text{m/s}^2 \). The negative sign indicates retardation (deceleration).

Step 5: Apply the Time Formula

Use the formula for time: \( t = \frac{v-u}{a} \). Substitute \( v = 0 \), \( u = 35 \, \text{m/s} \), and \( a = -3.0625 \, \text{m/s}^2 \).

Step 6: Calculate the Time Required

Calculate the time: \( t = \frac{0 - 35}{-3.0625} \approx \frac{-35}{-3.0625} \approx 11.4 \, \text{seconds} \).

Step 7: Verify with Options

Compare your results with the given options: \( a = 3.06 \, \text{m/s}^2 \) and \( t = 11.4 \, \text{seconds} \). So the closest correct answer is option (a).

Key Concepts

uniform motion

Uniform motion means moving at a steady speed without speeding up or slowing down. In our exercise, the car is initially moving at a constant speed before it starts to decelerate. This part of the movement characterizes uniform motion.
Understanding uniform motion is essential because it provides a baseline for what happens when an object does not experience any additional forces like braking or acceleration.
It's important to note that the speed during uniform motion remains consistent over time and is typically measured in units like kilometers per hour (km/h) or meters per second (m/s).

kinematic equations

Kinematic equations are fundamental when solving motion-related problems. They help connect different aspects like velocity, acceleration, time, and displacement. These equations assume uniform acceleration or deceleration, which means the rate of change of speed is constant.
The specific equation used in the exercise is:

• \[ v^2 = u^2 + 2as \]

Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration (or deceleration), and \(s\) is the distance covered. By rearranging this formula, we can solve for any missing variable.
This formula is particularly useful when you know the initial speed, the distance, and the final speed, as it allows you to find the car's retardation.

conversion of units

Conversion of units is an essential skill for solving physics problems accurately. Converting units ensures calculations remain consistent and results make sense in their given context. In this exercise, the initial speed is given in kilometers per hour (km/h), a common unit in everyday scenarios.
Calculating and predicting real-world scenarios often requires conversion to meters per second (m/s), which is a scientific unit. The conversion factor comes in handy here:

• \(1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s} \)
• Thus, \(126 \, \text{km/h} = \frac{126}{3.6} = 35 \, \text{m/s} \)

Accurate conversions are crucial for correctly applying kinematic equations and deriving the right solutions to problems. Without these conversions, the results would not be valid or useful when computing physics problems.

deceleration

Deceleration, also known as retardation, is the process of slowing down. It's the opposite of acceleration. In the exercise, the car slows down uniformly to a stop, illustrating this concept.
Deceleration is calculated using the formula

• \[ a = \frac{v^2 - u^2}{2s} \]

The negative sign in the final answer for acceleration indicates a decrease in speed over time, which is known as deceleration.
The resultant retardation value tells how quickly the car slows down over the stopping distance. It's typically measured in meters per second squared (m/s²). Understanding deceleration helps us study the effects of braking and the time it takes for an object to come to a complete stop.