/*! 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 44 A \(2-\mathrm{m}\) wide truck is... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 44

A \(2-\mathrm{m}\) wide truck is moving with a uniform speed \(v_{0}=8 \mathrm{~m} \mathrm{~s}^{-1}\) along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed \(v\) when the truck is \(4 \mathrm{~m}\) away from him. The minimum value of \(v\) so that he can cross the road safely is (1) \(2.62 \mathrm{~ms}^{-1}\) (2) \(4.6 \mathrm{~ms}^{-1}\) (3) \(3.57 \mathrm{~ms}^{-1}\) (4) \(1.414 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The minimum speed required is option (2) 4.6 ms鈦宦�.

Step by step solution

01

Calculate Time for Truck to Reach Pedestrian

The truck is initially 4 meters away from the pedestrian. Calculate the time it will take for the truck to reach the pedestrian using the formula for time, where time \( t = \frac{\text{distance}}{\text{rate}} \). The distance is 4 meters, and the speed of the truck is \( v_0 = 8\, \mathrm{ms}^{-1} \). Thus, \( t = \frac{4}{8} = 0.5 \) seconds.
02

Determine Time for Pedestrian to Cross Safely

The pedestrian needs to cover the width of the road, which is 2 meters, in the same amount of time or less. Therefore, the time available for the pedestrian to cross is the same as the truck reaching the initial position of the pedestrian: 0.5 seconds.
03

Calculate Minimum Speed Required for Pedestrian

Use the formula for speed \( v = \frac{\text{distance}}{\text{time}} \) to find the minimum speed of the pedestrian. The distance to be covered is the width of the road, 2 meters. The time is 0.5 seconds. Thus, \( v = \frac{2}{0.5} = 4 \mathrm{~ms}^{-1} \).
04

Compare with Given Options and Conclude

Among the given options, the closest value to 4 \( \mathrm{ms}^{-1} \) is option (2): \( 4.6 \mathrm{~ms}^{-1} \). This is the minimum speed with a small safety margin.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Uniform Speed
Uniform speed refers to the condition where an object moves at a constant rate, covering equal distances in equal intervals of time. When a moving body travels in such a way that its speed does not change over time, it is said to be in uniform motion. This means that the speed of the body is constant, and it does not speed up or slow down.
This concept is crucial in solving problems involving relative motion, especially when considering two objects moving at the same time. In the exercise, the truck moves at a uniform speed of 8 m/s, which helps us calculate how long it takes to reach a certain point. This uniformity simplifies calculations since acceleration doesn't need to be considered, making it a fundamental principle in introductory physics.
Time Calculation
Time calculation is integral in physics to determine how long an event or journey takes. It relies on the fundamental formula: \[t = \frac{d}{s}\]where:
  • \( t \) is the time,
  • \( d \) is the distance,
  • \( s \) is the speed.

For example, in the problem, knowing the uniform speed of the truck and the distance to the pedestrian, we calculate the time it takes the truck to reach the starting point of the pedestrian as 0.5 seconds. Accurate time calculation is essential, especially in safety considerations where timing can guide a pedestrian's safe crossing against approaching traffic.
Kinematics
Kinematics is the branch of mechanics that describes the motion of objects without considering the forces that cause the motion. It deals with concepts such as velocity, speed, acceleration, displacement, and time. Understanding kinematics is vital for solving the exercise as it allows us to analyze the relative motion between the truck and the pedestrian.

Equations of Motion

In kinematics, equations of motion apply to bodies moving in a straight line with uniform speed or uniform acceleration. The motion of both the truck and the pedestrian is a simple case of linear motion at constant velocities, underlining the kinematic principles used to solve the exercise.
Linear Motion
Linear motion, or rectilinear motion, refers to the movement of an object along a straight path. It is one-dimensional motion without any deviation in direction. This type of motion is effectively represented by parameters such as speed, distance, and time.

Practical Application

For the case study, both the truck and the pedestrian exhibit linear motion. The truck moves horizontally along the road, while the pedestrian walks straight across it. Understanding the straight-line motion simplifies analysis by reducing complexities involved with other types of movement such as curvilinear motion. By focusing on linear motion, it becomes easier to calculate time and speed requirements, ensuring accurate predictions of object positions over time.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The average velocity of a body moving with uniform acceleration after travelling a distance of \(3.06 \mathrm{~m}\) is \(0.34\) \(\mathrm{ms}^{-1} .\) If the change in velocity of the body is \(0.18 \mathrm{~m} \mathrm{~s}^{-1}\) during this time, its uniform acceleration is (1) \(0.01 \mathrm{~ms}^{-2}\) (2) \(0.02 \mathrm{~m} \mathrm{~s}^{-2}\) (3) \(0.03 \mathrm{~m} \mathrm{~s}^{-2}\) (4) \(0.04 \mathrm{~m} \mathrm{~s}^{-2}\)

From the top of a tower of height \(200 \mathrm{~m}\), a ball \(A\) is projected up with speed \(10 \mathrm{~ms}^{-1}\) and \(2 \mathrm{~s}\) later, another ball \(B\) is projected vertically down with the same speed. Then (1) Both \(A\) and \(B\) will reach the ground simultaneously (2) Ball \(A\) will hit the ground 2 s later than \(B\) hitting the ground (3) Both the balls will hit the ground with the same velocity (4) Both will rebound to the same height from the ground, if both have same coefficient of restitution.

A body is projected upwards with a velocity \(u\). It passes through a certain point above the ground after \(t_{1}\). The time afler which the body passes through the same point during the retum journey is (1) \(\left(\frac{u}{g}-t_{1}^{2}\right)\) (2) \(2\left(\frac{u}{g}-t_{1}\right)\) (3) \(3\left(\frac{u^{2}}{g}-t_{1}\right)\) (4) \(3\left(\frac{u^{2}}{g^{2}}-t_{1}\right)\)

A man swimming downstream overcomes a float at a point M. After travelling distance \(D\), he turned back and passed the float at a distance of \(D / 2\) from the point \(M\). Then the ratio of speed of swimmer with respect to still water to the speed of the river will be (1) 1 (2) 2 (3) 4 (4) 3

At time \(t=0\), a car moving along a straight line has a velocity of \(16 \mathrm{~ms}^{-1}\). It slows down with an acceleration of \(-0.5 t \mathrm{~m} \mathrm{~s}^{-2}\), where \(t\) is in seconds. Mark the correct statement(s). (1) The direction of velocity changes at \(t=8 \mathrm{~s}\). (2) The distance travelled in \(4 \mathrm{~s}\) is approximately \(59 \mathrm{~m}\). (3) The distance travelled by the particle in \(10 \mathrm{~s}\) is \(94 \mathrm{~m}\). (4) The velocity at \(t=10 \mathrm{~s}\) is \(9 \mathrm{~ms}^{-1}\)
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Famous Physicists

Read Explanation

Astrophysics

Read Explanation

Thermodynamics

Read Explanation

Modern Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.