/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 A motor car can be stopped withi... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 10

A motor car can be stopped within a distance of \(s\), when it moves with a speed \(v\). If it moves with a speed \(4 v\), it can be stopped within a distance (assuming constant braking force) (A) \(s\) (B) \(4 s\) (C) \(2 s\) (D) \(16 s\)

Short Answer

Expert verified
The stopping distance \(s'\), when a car moves at a speed of \(4v\), is \(16s\). Thus, the correct option is (D) \(16 s\).

Step by step solution

01

Recall the formula relating stopping distance, initial speed, and acceleration

We will use the formula: \(s = \frac{v^2}{2a}\), where \(s\) is the stopping distance, \(v\) is the initial speed, and \(a\) is the acceleration (which is negative due to braking force).
02

Apply the formula for the initial scenario

The initial scenario states that the car can be stopped within a distance of \(s\), when it moves with a speed \(v\). Thus, \(s = \frac{v^2}{2a}\).
03

Apply the formula for the modified scenario

Now, we need to apply the formula to calculate the stopping distance when the car moves with a speed of \(4v\). Let the new stopping distance be \(s'\). \(s' = \frac{(4v)^2}{2a}\).
04

Compare the stopping distances

Next, we need to compare the stopping distances of both scenarios. Our objective is to find the relationship between \(s'\) and \(s\). From Step 3, we have \(s' = \frac{(4v)^2}{2a} = \frac{16v^2}{2a}\). We can rewrite this expression in terms of \(s\), using the equation from Step 2: \(s = \frac{v^2}{2a}\). Thus, we have \(s' = 16 \times \frac{v^2}{2a} = 16s\).
05

Identify the correct answer

Comparing the stopping distances, we found that \(s' = 16s\). Therefore, the correct answer is: (D) \(16 s\)

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