/*! 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 111 The speed of a car increases uni... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 111

The speed of a car increases uniformly with time from \(50 \mathrm{km} / \mathrm{h}\) at \(A\) to \(100 \mathrm{km} / \mathrm{h}\) at \(B\) during \(10 \mathrm{sec}\) onds. The radius of curvature of the hump at \(A\) is \(40 \mathrm{m} .\) If the magnitude of the total acceleration of the mass center of the car is the same at \(B\) as at \(A,\) compute the radius of curvature \(\rho_{B}\) of the dip in the road at \(B\). The mass center of the car is \(0.6 \mathrm{m}\) from the road.

Short Answer

Expert verified
The radius of curvature at B is approximately 160 meters.

Step by step solution

01

Understanding the Problem

We need to find the radius of curvature \(\rho_B\) of a dip in the road at point \(B\) where the speed of the car is increased from \(50\, \mathrm{km/h}\) to \(100\, \mathrm{km/h}\) over \(10\, \mathrm{seconds}\). At the start point \(A\), we know that \(\rho_A = 40\, \mathrm{m}\). The magnitude of total acceleration at \(A\) and \(B\) remains the same.
02

Convert Units of Speed

Convert the initial and final speeds from km/h to m/s. We have:\[ v_A = 50\, \mathrm{km/h} = \frac{50\times1000}{3600}\, \mathrm{m/s} \approx 13.89\, \mathrm{m/s} \]\[ v_B = 100\, \mathrm{km/h} = \frac{100\times1000}{3600}\, \mathrm{m/s} \approx 27.78\, \mathrm{m/s} \]
03

Calculate Linear Acceleration

Using the formula for linear (tangential) acceleration, \( a_t = \frac{\Delta v}{\Delta t} \):\[ a_t = \frac{v_B - v_A}{t} = \frac{27.78 - 13.89}{10} \approx 1.389\, \mathrm{m/s^2} \]
04

Calculate Total Acceleration at A

The total acceleration \( a_A \) at point \(A\) is composed of tangential and centripetal components. The centripetal acceleration is given by \( a_c = \frac{v^2}{\rho} \). At \(A\), this becomes:\[ a_{cA} = \frac{(13.89)^2}{40} \approx 4.817\, \mathrm{m/s^2} \]Thus, the total acceleration \( a_A \) is:\[ a_A = \sqrt{a_t^2 + a_{cA}^2} = \sqrt{(1.389)^2 + (4.817)^2} \approx 5.000\, \mathrm{m/s^2} \]
05

Apply Condition at B

At \(B\), we must have the same total acceleration, \( a_B = a_A = 5.000\, \mathrm{m/s^2} \). Since \( a_B \) also consists of tangential and centripetal accelerations, we use:\[ a_{cB} = \sqrt{a_B^2 - a_t^2} = \sqrt{5.000^2 - 1.389^2} \approx 4.817\, \mathrm{m/s^2} \]
06

Solve for Radius of Curvature at B

Using the expression for centripetal acceleration \( a_{cB} = \frac{v_B^2}{\rho_B} \), we find \( \rho_B \) by rearranging:\[ \rho_B = \frac{(27.78)^2}{4.817} \approx 160\, \mathrm{m} \]

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.

Kinematics
Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause this motion. When analyzing kinematics, we focus on many different aspects like displacement, velocity, and acceleration.
  • Displacement is the change in position of an object, measured as a straight line between the starting and ending points.
  • Velocity is the rate of change of displacement, which tells us how fast an object is moving and what direction it is going.
  • Acceleration is the rate of change of velocity over a period of time.
In the exercise, the car鈥檚 speed changes over time, which is a key part of kinematics. By calculating how the speed changes, we can understand how the car's motion is evolving, such as when determining tangential acceleration.
To solve this particular problem, knowing the velocity at points A and B is essential in calculating the total acceleration experienced by the car.
Centripetal Acceleration
Centripetal acceleration occurs when an object is in circular motion, directed towards the center of the circular path. This acceleration is necessary for an object to change its direction while moving along the curve:
  • It is always perpendicular to the velocity of the object.
  • Can be calculated using the formula: \( a_c = \frac{v^2}{\rho} \), where \( v \) is the speed and \( \rho \) is the radius of curvature.
At point A, we calculated the car's centripetal acceleration to understand how it moves along the curvature of the hump. This acceleration combines with the tangential acceleration to give total acceleration.
Understanding centripetal acceleration is crucial because it reveals how much the path the car is taking is contributing to the overall experience of acceleration.
Radius of Curvature
The radius of curvature describes the radius of the circular path at any given point along a curve. This is crucial in circular motion:
  • The curve's sharpness can be inferred from the radius. A smaller radius means a sharper curve.
  • A larger radius of curvature indicates a gentler curve.
In the given exercise, finding the radius of curvature at point B, \( \rho_B \), is vital since it determines how the car maneuvers through the dip. By understanding the radius of curvature, we can predict the centripetal acceleration, which is necessary for solving problems of uniform circular motion.
Knowing the radius of curvature at different points allows engineers to design safer roads by predicting how much force will act on a vehicle.
Tangential Acceleration
Tangential acceleration is the component of acceleration that is parallel to the direction of motion. It affects how the speed of an object changes without altering its direction:
  • Calculated with \( a_t = \frac{\Delta v}{\Delta t} \), which presents how quickly the speed increases or decreases.
  • Acts along the direction of the object's path.
In the problem, we computed tangential acceleration to understand the rate at which the car accelerates along the curve from point A to point B. It is a direct reflection of the car鈥檚 increasing speed over the given time.
Understanding tangential acceleration helps in evaluating how fast a vehicle will speed up or slow down, contributing to both safety and performance assessments on curved paths.

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

When the effect of aerodynamic drag is included, the \(y\) -acceleration of a baseball moving vertically upward is \(a_{u}=-g-k v^{2},\) while the acceleration when the ball is moving downward is \(a_{d}=-g+k v^{2},\) where \(k\) is a positive constant and \(v\) is the speed in meters per second. If the ball is thrown upward at \(30 \mathrm{m} / \mathrm{s}\) from essentially ground level, compute its maximum height \(h\) and its speed \(v_{f}\) upon impact with the ground. Take \(k\) to be \(0.006 \mathrm{m}^{-1}\) and assume that \(g\) is constant.

At time \(t=0\) a small ball is projected from point \(A\) with a velocity of \(200 \mathrm{ft} / \mathrm{sec}\) at the \(60^{\circ}\) angle. Neglect atmospheric resistance and determine the two times \(t_{1}\) and \(t_{2}\) when the velocity of the ball makes an angle of \(45^{\circ}\) with the horizontal \(x\) -axis.

A train which is traveling at \(80 \mathrm{mi} / \mathrm{hr}\) applies its brakes as it reaches point \(A\) and slows down with a constant deceleration. Its decreased velocity is observed to be \(60 \mathrm{mi} / \mathrm{hr}\) as it passes a point \(1 / 2 \mathrm{mi}\) beyond \(A\). A car moving at \(50 \mathrm{mi} / \mathrm{hr}\) passes point \(B\) at the same instant that the train reaches point \(A\) In an unwise effort to beat the train to the crossing, the driver "steps on the gas." Calculate the constant acceleration \(a\) that the car must have in order to beat the train to the crossing by 4 seconds and find the velocity \(v\) of the car as it reaches the crossing.

The position \(s\) of a particle along a straight line is given by \(s=8 e^{-0.4 t}-6 t+t^{2},\) where \(s\) is in meters and \(t\) is the time in seconds. Determine the velocity \(v\) when the acceleration is \(3 \mathrm{m} / \mathrm{s}^{2}\)

The pilot of a jet transport brings the engines to full takeoff power before releasing the brakes as the aircraft is standing on the runway. The jet thrust remains constant, and the aircraft has a nearconstant acceleration of \(0.4 \mathrm{g}\). If the takeoff speed is \(200 \mathrm{km} / \mathrm{h},\) calculate the distance \(s\) and time \(t\) from rest to takeoff.
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Solid State Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Waves Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Work Energy and Power

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.