/*! 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 48 The speed limit in a school zone... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 48

The speed limit in a school zone is \(40 \mathrm{~km} / \mathrm{h}\) (about \(25 \mathrm{mi} / \mathrm{h}\) ). A driver traveling at this speed sees a child run onto the road \(13 \mathrm{~m}\) ahead of his car. He applies the brakes, and the car decelerates at a uniform rate of \(8.0 \mathrm{~m} / \mathrm{s}^{2} .\) If the driver's reaction time is \(0.25 \mathrm{~s},\) will the car stop before hitting the child?

Short Answer

Expert verified
No, the car needs a total distance of 15.89 m to stop, but only has 13 m available.

Step by step solution

01

Calculate Stopping Distance During Reaction Time

First, we need to determine how far the car travels during the driver's reaction time. The car's initial speed is given as \(40 \mathrm{~km/h}\), which we need to convert to meters per second by multiplying by \(\frac{1000}{3600}\). This gives us an initial speed \(v_i = \frac{40 \times 1000}{3600} = 11.11 \mathrm{~m/s}\). The distance covered during reaction time \(t_r = 0.25 \mathrm{~s}\) can be calculated using the formula \(d_r = v_i \times t_r\). Thus, \(d_r = 11.11 \times 0.25 = 2.78 \mathrm{~m}\).

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Key Concepts

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

Reaction Time
When driving, reaction time is the period between recognizing a hazard and taking action, like pressing the brakes. In this context, the driver sees a child ahead and takes a moment to react. Even though it seems instant, this time delay can influence the stopping distance significantly. Imagine the driver traveling at a speed of 40 km/h. During those precious 0.25 seconds of reaction, the car continues to move.
  • This delay results in the car traveling a certain distance forward, despite the driver's intention to stop.
  • Understanding reaction time emphasizes the need for maintaining reasonable speed and distance in hazardous situations, particularly in school zones where children might suddenly appear.
Deceleration
Deceleration is the rate at which a car slows down. In our example, deceleration is uniform at 8.0 m/s虏. This means the car consistently reduces its speed by 8.0 m/s for every second it's braking. The objective is to calculate if this rate allows the vehicle to stop in time. To understand deceleration, picture the car's speed reducing at consistent intervals:
  • A high rate of deceleration means the car can stop quickly.
  • This ability is crucial to avoid collisions, especially in unpredictable scenarios.
By knowing the exact deceleration rate and the initial speed, one can determine how much distance the car will cover before it comes to a complete stop.
Speed Conversion
In the realm of physics and mathematics, speed conversion allows us to switch between different units of measurement. For instance, our given speed of 40 km/h must be converted to m/s for accurate calculations within this problem context. This is achieved by:
  • First recognizing that 1 kilometer equals 1000 meters, and 1 hour equals 3600 seconds.
  • Thus, converting the speed involves multiplying by \( \frac{1000}{3600} \), resulting in an initial speed of 11.11 m/s.
Understanding how to perform speed conversions ensures that all units are consistent, improving calculation accuracy. This is crucial because any discrepancy in units could lead to significant errors, particularly in safety-critical situations like stopping a vehicle in time.

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Most popular questions from this chapter

A train normally travels at a uniform speed of \(72 \mathrm{~km} / \mathrm{h}\) on a long stretch of straight, level track. On a particular day, the train must make a 2.0 -min stop at a station along this track. If the train decelerates at a uniform rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) and, after the stop, accelerates at a rate of \(0.50 \mathrm{~m} / \mathrm{s}^{2},\) how much time is lost because of stopping at the station?

A hospital nurse walks \(25 \mathrm{~m}\) to a patient's room at the end of the hall in 0.50 min. She talks with the patient for 4.0 min, and then walks back to the nursing station at the same rate she came. What was the nurse's average speed?

Two runners approaching each other on a straight track have constant speeds of \(4.50 \mathrm{~m} / \mathrm{s}\) and \(3.50 \mathrm{~m} / \mathrm{s}\) respectively, when they are \(100 \mathrm{~m}\) apart ( -Fig. 2.22). How long will it take for the runners to meet, and at what position will they meet if they maintain these speeds?

At a sports car rally, a car starting from rest accelerates uniformly at a rate of \(9.0 \mathrm{~m} / \mathrm{s}^{2}\) over a straight-line distance of \(100 \mathrm{~m}\). The time to beat in this event is \(4.5 \mathrm{~s}\). Does the driver beat this time? If not, what must the minimum acceleration be to do so?

A motorboat traveling on a straight course slows uniformly from \(60 \mathrm{~km} / \mathrm{h}\) to \(40 \mathrm{~km} / \mathrm{h}\) in a distance of \(50 \mathrm{~m}\) What is the boat's acceleration?
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