/*! 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 147 A car is moving with speed \(30 ... [FREE SOLUTION] | 91Ó°ÊÓ

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A car is moving with speed \(30 \mathrm{~m} / \mathrm{sec}\) on a circular path of radius \(500 \mathrm{~m}\). Its speed is increasing at the rate of \(2 \mathrm{~m} / \mathrm{sec}^{2} .\) What is the acceleration of the car \(\quad\) [Roorkee 1982; RPET 1996; MH CET 2002; MP PMT 2003] (a) \(2 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(2.7 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(9.8 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The acceleration of the car is \(2.7 \mathrm{~m} / \mathrm{s}^{2}\), so the correct answer is (b) \(2.7 \mathrm{~m} / \mathrm{s}^{2}\).

Step by step solution

01

- Determine Given Values

Identify the given values. From the question, we have the speed of the car (v) as 30 m/s, the rate at which speed is increasing (a1) as 2 m/s^2, and the radius of the circular path (r) as 500 m.
02

- Calculating Centripetal Acceleration

Calculate the centripetal (or radial) acceleration using the formula, a_centripetal = \(v^2 / r\). Substituting the given values, we get a_centripetal = \( (30 m/s)^2 / 500 m = 1.8 m/s^2 \).
03

Calculate Total Acceleration

Compute the total acceleration using the formula \( a = \sqrt{a_{tangential}^2 + a_{centripetal}^2} \), where a_tangential is the tangential acceleration (the rate at which speed is increasing) of 2 m/s^2 and a_centripetal is the acceleration calculated in Step 2. Substituting the values, we get a = \( \sqrt{(2 m/s^2)^2 + (1.8 m/s^2)^2} = 2.7 m/s^2 \).

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Key Concepts

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

Centripetal Acceleration
When an object moves in a circle, it constantly changes direction which requires an acceleration. This inward acceleration is known as centripetal acceleration. The formula to calculate centripetal acceleration is given by \(a_{\text{centripetal}} = \frac{v^2}{r}\), where \(v\) is the velocity of the object and \(r\) is the radius of the circular path.
For example, if a car travels at a speed of 30 m/s on a circular track with a radius of 500 m, substitute these values into the formula to find that the centripetal acceleration \(a_{\text{centripetal}} = \frac{(30 \text{ m/s})^2}{500 \text{ m}} = 1.8 \text{ m/s}^2\).
This acceleration ensures the car maintains its circular path by continuously pulling it inward toward the center of the circle.
  • Always acts towards the center of the circle
  • It doesn’t change the speed, only the direction
  • Proportional to the square of the speed
Understanding this concept is crucial in circular motion as it explains how objects can sustain circular tracks without veering off.
Tangential Acceleration
Tangential acceleration refers to the change in speed of an object moving along a circular path. It's the component of acceleration in the direction of the tangent to the path at any point.
If you think of a car increasing its speed as it drives, it experiences tangential acceleration. In the exercise, the car's speed increases at a rate of 2 m/s², meaning the car gains an additional 2 meters per second every second in speed.
  • Affects the speed of the object
  • Occurs along the direction of motion
  • Different from centripetal acceleration which affects the direction
This acceleration, combined with centripetal acceleration, gives us the object's total acceleration.
Understanding tangential acceleration is vital to grasp how forces work on objects moving in a non-uniform circular motion, allowing them to speed up or slow down.
Total Acceleration
Total acceleration in circular motion is a combination of both centripetal and tangential accelerations. Both components are perpendicular to each other. When both are present, the total acceleration can be computed using the Pythagorean theorem.
In this case, we use the formula \(a = \sqrt{a_{\text{tangential}}^2 + a_{\text{centripetal}}^2}\), where \(a_{\text{tangential}}\) is 2 m/s² and \(a_{\text{centripetal}}\) is 1.8 m/s². Plugging these into the formula gets \(a = \sqrt{(2 \text{ m/s}^2)^2 + (1.8 \text{ m/s}^2)^2} = 2.7 \text{ m/s}^2\).
  • Represents the overall change in velocity
  • Affects both speed and direction
  • Important in non-uniform circular motion
Thus, total acceleration provides a holistic view of how an object’s velocity vector changes in motion and is essential for analyzing movement dynamics in circular paths.

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Most popular questions from this chapter

3\. If a cyclist moving with a speed of \(4.9 \mathrm{~m} / \mathrm{s}\) on a level road can take a sharp circular turn of radius \(4 \mathrm{~m}\), then coefficient of friction between the cycle tyres and road is (a) \(0.41\) (b) \(0.51\) (c) \(0.71\) (d) \(0.61\)

A particle is moving along a circular path of radius 3 meter in such a way that the distance travelled measured along the circumference is given by \(S=\frac{t^{2}}{2}+\frac{t^{3}}{3} .\) The acceleration of particle when \(t=2 \mathrm{sec}\) is (a) \(1.3 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(13 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(3 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(10 \mathrm{~m} / \mathrm{s}^{2}\)

A proton of mass \(1.6 \times 10^{-27} \mathrm{~kg}\) goes round in a circular orbit of radius \(0.10 \mathrm{~m}\) under a centripetal force of \(4 \times 10^{-13} N .\) then the frequency of revolution of the proton is about [Kerala PMT 2002] (a) \(0.08 \times 10^{8}\) cycles per sec (b) \(4 \times 10^{8}\) cycles per sec (c) \(8 \times 10^{8}\) cycles per sec (d) \(12 \times 10^{8}\) cycles per sec

An athlete completes one round of a circular track of radius \(10 \mathrm{~m}\) in \(40 \mathrm{sec}\). The distance covered by him in \(2 \min 20\) sec is [Kerala PMT 2002] (a) \(70 \mathrm{~m}\) (b) \(140 \mathrm{~m}\) (c) \(110 \mathrm{~m}\) (d) \(220 \mathrm{~m}\)

Keeping the banking angle same to increase the maximum speed with which a car can travel on a curved road by \(10 \%\), the radius of curvature of road has to be changed from \(20 \mathrm{~m}\) to (a) \(16 m\) (b) \(18 m\) (c) \(24.25 \mathrm{~m}\) (d) \(30.5 \mathrm{~m}\)

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