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Chapter 3: Problem 81

A motor car is traveling at \(60 \mathrm{~m} / \mathrm{s}\) on a circular road of radius \(1200 \mathrm{~m}\). It is increasing its speed at the rate of \(4 \mathrm{~m} / \mathrm{s}^{2}\). The acceleration of the car is (A) \(3 \mathrm{~ms}^{-2}\) (B) \(4 \mathrm{~ms}^{-2}\) (C) \(5 \mathrm{~ms}^{-2}\) (D) \(7 \mathrm{~ms}^{-2}\)

Short Answer

Expert verified
The short answer based on the step-by-step solution is: The total acceleration of the car is \(5 \mathrm{~m/s^2}\), which corresponds to option (C).

Step by step solution

01

Write down the given values.

We know the following: - Linear speed, \(v = 60 \mathrm{~m/s}\) - Radius of circular motion, \(r = 1200 \mathrm{~m}\) - Tangential acceleration, \(a_t = 4 \mathrm{~m/s^2}\)
02

Calculate centripetal acceleration.

Centripetal acceleration is the acceleration towards the center of the circular path, and can be calculated using the formula: \(a_c = \frac{v^2}{r}\) Substitute the known values into the formula: \(a_c = \frac{(60 \mathrm{~m/s})^2}{1200 \mathrm{~m}}\) Calculate the result: \(a_c = 3 \mathrm{~m/s^2}\)
03

Determine the total acceleration.

Since the centripetal and tangential accelerations are perpendicular to each other, we can find the total acceleration by applying the Pythagorean theorem: Total acceleration, \(a = \sqrt{a_c^2 + a_t^2}\) Substitute the known values into the formula: \(a = \sqrt{(3 \mathrm{~m/s^2})^2 + (4 \mathrm{~m/s^2})^2}\) Calculate the result: \(a = \sqrt{9 + 16}\) \(a = \sqrt{25}\) \(a = 5 \mathrm{~m/s^2}\)
04

Select the correct answer.

The total acceleration of the car is \(5 \mathrm{~m/s^2}\), which corresponds to option (C). Therefore, the correct answer is: (C) \(5 \mathrm{~ms}^{-2}\)

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Most popular questions from this chapter

A fireman wants to slide down a rope. The breaking load for the rope is \(\frac{3}{4}\) th of the weight of the fireman. The acceleration of the fireman to prevent the rope from breaking will be (Acceleration due to gravity is \(g\) ) (A) \(g / 4\) (B) \(g / 2\) (C) \(3 g / 4\) (D) Zero

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