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Chapter 1: Problem 60

A particle has an acceleration of \(72 \mathrm{~km} / \mathrm{min}^{2}\) find acceleration in SI system. (a) \(0.5 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(30 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(18 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(20 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The correct answer is (d) \(20 \mathrm{~m/s}^{2}\).

Step by step solution

01

Identify the given acceleration in km/min²

The given acceleration is 72 km/min².
02

Convert km to meters

We know that 1 km = 1000 m. To convert the given acceleration from km to meters, we'll multiply by 1000. \(72 \mathrm{~km} \times 1000 = 72000 \mathrm{~m}\) So, the acceleration is now \(72000 \mathrm{~m/min}^{2}\).
03

Convert minutes to seconds

We know that 1 minute = 60 seconds. To convert acceleration from m/min² to m/s², we'll divide by the square of 60. (As the time is squared) \(72000 \mathrm{~m/min}^{2} \div (60 \times 60) = 72000 \mathrm{~m/min}^{2} \div 3600 = 20 \mathrm{~m/s}^{2}\)
04

Find the correct option

Now that we have found the acceleration in the SI system, which is \(20 \mathrm{~m/s}^{2}\), we can see that the correct option is: (d) \(20 \mathrm{~m/s}^{2}\)

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