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Chapter 3: Problem 67

The distance moved by motor cycle, \(\mathrm{d}=10267-10237\) \(=30 \mathrm{~km}\) The time taken, \(\mathrm{t}=9: 00 \mathrm{am}-8: 30 \mathrm{am}=30 \mathrm{~min}\) \(=\frac{1}{2} \mathrm{~h}\) The speed of motor cycles \(\mathrm{v}=\frac{\mathrm{d}}{\mathrm{t}}=\frac{30 \mathrm{~km}}{\frac{1}{2} \mathrm{~h}}=60 \mathrm{~km} \mathrm{~h}^{-1}\).

Short Answer

Expert verified
Answer: The speed of the motorcycle is 60 km/h.

Step by step solution

01

Find the distance moved by the motorcycle

To find the distance moved, subtract the initial position from the final position: \(d = 10267 - 10237 = 30 \mathrm{~km}\)
02

Calculate the time taken to move the distance

Convert the given times into minutes and find the difference between them. Then, convert that difference back to hours: \(30\) min \(= \frac{1}{2}\) hour
03

Calculate the speed of the motorcycle

Now, we can find the speed by dividing the distance by time: \(v = \frac{d}{t} = \frac{30 \mathrm{~km}}{\frac{1}{2} \mathrm{~h}} = 60 \mathrm{~km} \mathrm{~h}^{-1}\) The speed of the motorcycle is \(60 \mathrm{~km} \mathrm{~h}^{-1}\).

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Most popular questions from this chapter

The time period of ' \(\mathrm{A}\) ' is, \(\mathrm{T}_{1}=\frac{\mathrm{T}}{20}\). The time period of ' \(\mathrm{B}\) ' is, \(\mathrm{T}_{2}=\frac{\mathrm{T}}{30}\) The ratio of time periods of ' \(A\) ' and ' \(B\) ' is \(\frac{T_{1}}{T_{2}}=\frac{\frac{T}{20}}{\frac{T}{30}}=\frac{30}{20}=\frac{3}{2}\) \(T_{1}: T_{2}=3: 2\)

The number of oscillations completed in one second is called its frequency. The bob is oscillating once in every 6 seconds The number of oscillations in \(1 \mathrm{~s}=1 / 6\)

Unit of time in SI and CGS systems is second (s). Unit of speed in SI and CGS systems are \(\mathrm{m} \mathrm{s}^{-1}\) and \(\mathrm{cm} \mathrm{s}^{-1}\) respectively.

\(1 \mathrm{~km} \mathrm{~h}^{-1}=10^{3} \mathrm{~m} \times \mathrm{h}^{-1}\) \(=\frac{1000 \mathrm{~m}}{1 \mathrm{~h}}=\frac{1000 \mathrm{~m}}{60 \times 60 \mathrm{~s}}\) \(=\frac{10}{36} \mathrm{~m} \mathrm{~s}^{-1}=\frac{5}{18} \mathrm{~m} \mathrm{~s}^{-1}\)

(i) The unit of speed and velocity is \(\mathrm{m} \mathrm{s}^{-1}\). (ii) If a body is moving in a straight line path then its speed is equal to velocity \(50 \mathrm{~m} \mathrm{~s}^{-1}=50 \times \frac{18}{5} \mathrm{~km} \mathrm{~h}^{-1}=180 \mathrm{~km} \mathrm{~h}^{-1}\)
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