Problem 81 - A Complete Resource Book in Physics for JEE Main

Question

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) \(7 \mathrm{~ms}^{-2}\)

Step-by-Step Solution

Answer

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 1: 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, \(a_t = 4 \mathrm{~m/s^2}\)

Step 2: 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: \(a_c = 3 \mathrm{~m/s^2}\)

Step 3: 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: \(a = \sqrt{9 + 16}\) \(a = \sqrt{25}\) \(a = 5 \mathrm{~m/s^2}\)

Step 4: 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) \(5 \mathrm{~ms}^{-2}\)