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Chapter 2: Problem 183

A car moving over a straight path covers a distance \(\mathrm{x}\) with constant speed \(10 \mathrm{~ms}^{-1}\) and then the same distance with constant speed of \(\mathrm{V}_{2}\). If average speed of the car is \(16 \mathrm{~ms}^{-1}\), then \(\mathrm{V}_{2}=\ldots\) (A) \(30 \mathrm{~ms}^{-1}\) (B) \(20 \mathrm{~ms}^{-1}\) (C) \(40 \mathrm{~ms}^{-1}\) (D) \(25 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The value of \(V_2\) is \(40ms^{-1}\). So, the correct answer is (C).

Step by step solution

01

Write down the given values and formula

We have the following given values: - Initial speed (\(V_1\)): \(10ms^{-1}\) - Second speed (\(V_2\)): unknown - Average speed (\(V_{avg}\)): \(16ms^{-1}\) - Distance x: same for both speeds The formula for average speed is: \[V_{avg}=\dfrac{\text{Total Distance}}{\text{Total Time}}\] Since the car travels the same distance x twice, Total distance = \(2x\).
02

Find the time taken during each part of the journey

To find the total time, we need to find the time it takes for the car to travel each part of the journey. For the first part of the journey: \[t_1=\dfrac{x}{V_1}\] For the second part of the journey: \[t_2=\dfrac{x}{V_2}\]
03

Calculate the total time

To extract the total time, we just sum the previously found \(t_1\) and \(t_2\): \[t_{total} = t_{1}+t_{2}=\dfrac{x}{10}+\dfrac{x}{V_2}\]
04

Use the formula for average speed and given values to find \(V_2\)

We can now plug in the values for \(V_{avg}\), Total Distance and Total Time into the average speed formula: \[16 = \dfrac{2x}{\frac{x}{10}+\frac{x}{V_2}}\] Now, we need to solve for \(V_2\).
05

Cross multiply and arrange the equation.

Cross-multiply to eliminate the fraction in the equation: \[16\left(\frac{x}{10}+\frac{x}{V_2}\right) = 2x\] Then, distribute 16: \[1.6x+\frac{16x}{V_2} = 2x\] Now, isolate terms with \(V_2\) and rearrange the equation: \[\frac{16x}{V_2} = 2x - 1.6x\Longrightarrow \frac{16x}{V_2} = 0.4x\]
06

Solve for \(V_2\)

Now, cross multiply to eliminate the remaining fraction: \(16x = 0.4xV_2\) Now, divide by 0.4x to get \(V_2\): \(\dfrac{16x}{0.4x} = V_2\) Finally, divide 16 by 0.4: \(V_2 = \dfrac{16}{0.4} = 40ms^{-1}\) So, the correct answer is: (C) \(40ms^{-1}\).

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