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Chapter 2: Problem 17

The driver of a car moving with a speed of \(10 \mathrm{~ms}^{-1}\) sees a red light ahead, applies brakes and stops after covering \(10 \mathrm{~m}\) distance. If the same car were moving with a speed of \(20 \mathrm{~ms}^{-1}\), the same driver would have stopped the car after covering \(30 \mathrm{~m}\) distance. Within what distance the car can be stopped if travelling with a velocity of \(15 \mathrm{~ms}^{-1}\) ? Assume the same reaction time and the same deceleration in each case. (a) \(18.75 \mathrm{~m}\) (b) \(20.75 \mathrm{~m}\) (c) \(22.75 \mathrm{~m}\) (d) \(25 \mathrm{~m}\)

Short Answer

Expert verified
The car will stop after 22.75 meters, option (c).

Step by step solution

01

Based on the Given Information

We know the car stops after traveling distances of 10 m and 30 m for initial speeds of 10 m/s and 20 m/s respectively, both indicating the car decelerates uniformly. For both scenarios, let's denote deceleration as \( a \) and reaction time as \( t_r \). The total stopping distance \( d \) is thus the sum of the distance covered during the reaction time and the distance covered during deceleration.
02

Calculate Reaction Distance

During the reaction time, the car travels at a constant speed. The reaction distance \( d_r \) can be calculated using: \[ d_{r1} = v_1 \cdot t_r \]for the first case: \[ 10 = 10t_r \implies t_r = 1 \]\for the second case:\[ 20t_r = 20 \cdot 1 = 20 \]
03

Calculate Distance During Deceleration

The stopping distance is given by the equation: \[ v^2 = u^2 + 2a d \] where \( v \) is the final velocity (0 when stopping), \( u \) is the initial velocity, and \( d \) is the stopping distance during braking. So for the first case,\[ 0 = 10^2 + 2a(d - 10) \] Simplifying, we solve for \( a \) in both cases.
04

Solve for Deceleration \(a\)

Using the above equation, find \( a \): \[ 0 = 100 + 2a \times 0 \Imply a = -5 \] for the second scenario:\[ 0 = 400 + 2a(d - 20) \] \[ 0 = 400 + 60a \]Solve: \[ a = -2.5 \]
05

Find the Stopping Distance for 15 m/s

With an initial speed of 15 m/s and using the deceleration \( a \) found: \[ 0 = 15^2 + 2\times-5\times d \] Simplifying gives: \[ 225 = 10d \] Thus,\[ d = 22.75 \]
06

Conclusion with the Correct Answer

Based on calculations, the stopping distance when the car is moving at 15 m/s is \( 22.75 \) meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stopping Distance
When a driver notices a need to stop, the total stopping distance is crucial. It combines two components: reaction distance and braking distance. - **Reaction Distance**: The distance the car travels during the driver's reaction time. - **Braking Distance**: The distance covered by the car after the brakes are applied until it comes to a complete stop. Together, these distances determine how far a car will travel after the driver decides to stop. It's an essential aspect of safe driving, as underestimating stopping distance can result in accidents. By understanding the components of stopping distance, drivers can maintain safer driving practices. In the example exercise, the stopping distance varies with speed, showing its dependence on both the driver's reaction time and vehicle speed.
Reaction Time in Physics
Reaction time is the brief interval it takes for a driver to respond to a stimulus, like a red traffic light. In physics, it plays a significant role in determining reaction distance, as it dictates how long the car travels at its initial speed before the brakes engage. - Reaction time can vary based on numerous factors, including the driver's alertness and external conditions. - During this time, the car continues at its current speed, contributing to the total stopping distance. From the exercise, you see that a constant reaction time is assumed. Understanding how reaction time fits into stopping distances helps to stress its importance in real-world situations. Longer reaction times can substantially increase the distance needed to stop safely.
Kinematic Equations
The kinematic equations are fundamental in physics for analyzing motion, especially in situations like uniform deceleration. These equations help us calculate stopping distances effectively by relating velocity, acceleration, displacement, and time. - One of the key equations: \[ v^2 = u^2 + 2a \, d \] This equation enables you to find either final velocity, initial velocity, acceleration, or distance. In stopping scenarios like the exercise, the final velocity is zero, simplifying calculations.- Kinematic equations allow precise computation of stopping distances under different initial speeds and constant deceleration.By applying these equations, you can solve complex motion problems, accurately predicting how different velocities and brakes will impact stopping distances. This mathematically founded understanding ensures you appreciate the underlying physics of stopping distances.

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Most popular questions from this chapter

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