/*! 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 3 A car moving along a circular tr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 3

A car moving along a circular track of radius \(50.0 \mathrm{~m}\) accelerates from rest at \(3.00 \mathrm{~ms}^{-2}\). Consider a situation when the car's centripetal acceleration equals its tangential acceleration. Then (1) The angle around the track does the car travel is 1 rad (2) The magnitude of the car's total acceleration at that instant is \(3 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\) (3) Time elapses before this situation is \(\sqrt{\frac{50}{3}} \mathrm{~s}\). (4) The distance travelled by the car during this time is \(25 \mathrm{~m}\)

Short Answer

Expert verified
Options (1), (2), (3) are correct; (4) should be 25 m.

Step by step solution

01

Understanding the Problem

A car is accelerating along a circular track with both tangential and centripetal acceleration components. The problem asks for different quantities when these two accelerations are equal.
02

Centripetal and Tangential Accelerations

Centripetal acceleration formula is given by \(a_c = \frac{v^2}{r}\). Tangential acceleration here is \(a_t = 3.00 \mathrm{~ms}^{-2}\). Setting \(a_c = a_t\), we get \(\frac{v^2}{r} = 3.00 \).
03

Solving for Velocity

With \(r = 50.0 \mathrm{~m}\), substitute it into the equation \(\frac{v^2}{50} = 3.00\). Solving for \(v\), we get \(v^2 = 150\). Therefore, \(v = \sqrt{150}\).
04

Find Total Acceleration

Total acceleration \(a\) in terms of components is \(a = \sqrt{a_t^2 + a_c^2}\). Since \(a_t = a_c = 3.00\), thus \(a = \sqrt{3^2 + 3^2} = 3\sqrt{2} \mathrm{~ms}^{-2}\).
05

Determine Time Elapsed

Since \(v = u + a_t t\) (where initial velocity \(u = 0\) because the car starts from rest), \(v = 0 + 3t = \sqrt{150}\), solving gives \(t = \sqrt{\frac{50}{3}}\).
06

Calculate Distance Traveled

Distance \(s\) can be found using \(s = ut + \frac{1}{2} a_t t^2\). Here \(u = 0\) gives \(s = \frac{1}{2} \times 3 \times \left(\sqrt{\frac{50}{3}} \right)^2 = 25 \mathrm{~m}\).
07

Angle Traveled Around the Track

The angle \(\theta\) in radians is \(\theta = \frac{s}{r}\). Given \(s = 25\mathrm{~m}\) and \(r = 50 \mathrm{~m}\), \(\theta = \frac{25}{50} = 0.5 \mathrm{~rad}\).

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.

Circular Motion
Circular motion occurs when an object moves along a circular path. It implies a continuous change in direction, thus a change in velocity, even if the speed remains constant.
In this exercise, a car travels along a circular track, indicating circular motion. When objects travel in a circle, two types of accelerations affect them:
- **Centripetal Acceleration**, which is directed towards the center of the circle. It is necessary to keep the car on its curved path and given by the formula: \[ a_c = \frac{v^2}{r} \]where \(a_c\) is the centripetal acceleration, \(v\) is the velocity, and \(r\) is the radius of the circle.
- **Tangential Acceleration**, which changes the object's speed in its path along the perimeter of the circle. It's analogous to acceleration on a straight path. Together, these accelerations define the motion characteristics of the car around the circular track.
Acceleration Components
Acceleration is the rate of change of velocity. In circular motion, it can be broken down into two components: radial (centripetal) and tangential.
  • **Centripetal Component**: Keeps the car moving along the circular path. Without it, the car would continue in a straight line. It acts perpendicular to the tangential velocity and depends on the velocity of the car and the radius of the circle.
  • **Tangential Component**: Responsible for changing the speed of the car as it moves around the circle. It is aligned with the motion of the car and in this problem, it is given as \(3.00 \mathrm{~ms}^{-2}\).
When centripetal and tangential accelerations are equal, as in this problem, it implies that the speed and direction are changing at the same rate. This special scenario allows for direct comparison and calculation of various motion parameters, such as time, speed, and distance.
Physics Problems for JEE Advanced
Physics problems in competitive exams like JEE Advanced often involve multi-step solutions and require a good understanding of several concepts at once. This problem is a classic example:
- It involves understanding both tangential and centripetal acceleration and their interplay in circular motion.
- It requires solving equations to find unknowns like time and distance using given acceleration values.

To solve successfully, students need to
  • Break down the problem into simpler parts.
  • Use appropriate formulas to relate quantities.
  • Understand concepts of simultaneous equalization (as when \(a_t = a_c\)).
By following this systematic approach, complex physics problems become more manageable and intuitive, aiding in mastery for the JEE Advanced exam.

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

Ball \(A\) of mass \(m\), after sliding from an inclined plane, strikes elastically another ball \(B\) of same mass at rest. Find the minimum height \(h\) so that ball \(B\) just completes the circular motion of the surface at \(C\). (All surfaces are smooth.) (1) \(h=\frac{5}{2} R\) (2) \(h=2 R\) (3) \(h=\frac{2}{5} R\) (4) \(h=3 R\)

\(A\) and \(B\) are moving in 2 circular orbits with angular velocity \(2 \omega\) and \(\omega\) respectively. Their positions are as shown at \(t=0 .\) Find the time when they will meet for first time. (1) \(\frac{\pi}{2 \omega}\) (2) \(\frac{3 \pi}{2 \omega}\) (3) \(\frac{\pi}{\omega}\) (4) they will never meet

A particle of mass \(m\) moves along a circular path of radius \(r\) with a centripetal acceleration \(a_{n}\) changing with time \(t\) as \(a_{n}=k t^{2}\), where \(k\) is a positive constant. The average power developed by all the forces acting on the particle during the first \(t_{0}\) seconds is (1) \(m k r t_{0}\) (2) \(\frac{m k r t_{0}^{2}}{2}\) (3) \(\frac{m k r t_{0}}{2}\) (4) \(\frac{m k r t_{0}}{4}\)

The track of motorcycle-race is circular and unbanked. There are two bikers on the road, one travels along a path of greater radius than the other. They both lean towards the centre at the same angle. Which one completes the circular path in less time? (1) biker having path of smaller radius. (2) biker having path of larger radius. (3) will complete circle in same time. (4) both can not lean at same angle if radius is different.

Figure shows a smooth vertical circular track \(A B\) of radius \(R\). A block slides along the surface \(A B\) when it is given a velocity equal to \(\sqrt{6 g R}\) at point \(A\). The ratio of the force exerted by the track on the block at point \(A\) to that at point \(B\) is (1) \(0.25\) (2) \(0.35\) (3) \(0.45\) (-1) \(0.55\)
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Atoms and Radioactivity

Read Explanation

Fluids

Read Explanation

Dynamics

Read Explanation

Electricity and Magnetism

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.