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Chapter 2: Problem 185

A car covers one third part of its straight path with speed \(\mathrm{V}_{1}\) and the rest with speed \(\mathrm{V}_{2}\). What is its average speed? (A) $\left[\left(3 \mathrm{v}_{1} \mathrm{v}_{2}\right) /\left(2 \mathrm{v}_{1}+\mathrm{v}_{2}\right)\right]$ (B) $\left[\left(2 \mathrm{v}_{1} \mathrm{v}_{2}\right) /\left(3 \mathrm{v}_{1}+\mathrm{v}_{2}\right)\right]$ (C) $\left[\left(3 \mathrm{v}_{1} \mathrm{v}_{2}\right) /\left(\mathrm{v}_{1}+2 \mathrm{v}_{2}\right)\right]$ (D) $\left[\left(3 \mathrm{v}_{1} \mathrm{v}_{2}\right) /\left(2 \mathrm{v}_{1}+2 \mathrm{v}_{2}\right)\right]$

Short Answer

Expert verified
The short answer is: \(\boxed{\text{(D)}\:\frac{3V_1V_2}{2V_1+2V_2}}\).

Step by step solution

01

Calculate the time taken for each part

Let \(d_1\) be the distance covered with speed \(V_1\), and \(d_2\) be the distance covered with speed \(V_2\). We know that \(d_1 = \frac{1}{3}d\) (where \(d\) is the total distance), and \(d_2 = \frac{2}{3}d\). Now, using the formula of time, \[Time = \frac{Distance}{Speed}\] Calculate the time taken for each part: \[t_1 = \frac{d_1}{V_1} = \frac{\frac{1}{3}d}{V_1}\] \[t_2 = \frac{d_2}{V_2} = \frac{\frac{2}{3}d}{V_2}\]
02

Calculate Total Time

Add the time taken for each part to calculate the total time: \[Total\:Time = t_1 + t_2 = \frac{\frac{1}{3}d}{V_1} + \frac{\frac{2}{3}d}{V_2}\]
03

Calculate the Average Speed

Now, using the formula for the average speed and the total time calculated in step 2, \[Average\:Speed = \frac{Total\:Distance}{Total\:Time} = \frac{d}{\frac{\frac{1}{3}d}{V_1} + \frac{\frac{2}{3}d}{V_2}}\]
04

Simplify the formula and compare to the given options

\[Average\:Speed = \frac{d}{\frac{\frac{1}{3}d}{V_1} + \frac{\frac{2}{3}d}{V_2}} \cdot \frac{3V_1V_2}{3V_1V_2}\] \[Average\:Speed = \frac{3V_1V_2d}{\left(\frac{1}{3}dV_2\right) + \left(\frac{2}{3}dV_1\right)}\] \[Average\:Speed = \frac{3V_1V_2d}{\left(V_2d\right) + \left(2V_1d\right)}\] Since the total distance (d) is the same for both parts, we can cancel it out: \[Average\:Speed = \frac{3V_1V_2}{\left(V_2\right) + \left(2V_1\right)}\] Comparing to the given options, the final answer is (D) \(\left[\left(3 \mathrm{v}_{1} \mathrm{v}_{2}\right) /\left(2 \mathrm{v}_{1}+2 \mathrm{v}_{2}\right)\right]\).

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Most popular questions from this chapter

An object is projected with initial velocity of \(100 \mathrm{~ms}^{-1}\) and angle of 60 . Find the vertical velocity when its horizontal displacement is \(500 \mathrm{~m} .\left(\mathrm{g}=10 \mathrm{~ms}^{-1}\right)\) (A) \(13.35 \mathrm{~ms}^{-1}\) (B) \(-13.35 \mathrm{~ms}^{-1}\) (C) \(-8.65 \mathrm{~ms}^{-1}\) (D) \(98 \mathrm{~ms}^{-1}\)

A particle in \(\mathrm{xy}\) plane is governed by $\mathrm{x}=\mathrm{A} \cos \omega \mathrm{t}, \mathrm{y}=\mathrm{A}$ \((1-\sin \omega \mathrm{t}) . \mathrm{A}\) and \(\omega\) are constants. What is the speed of the particle. (A) A\omegat (B) \(\mathrm{A} \omega^{2} \mathrm{t}\) (C) \(\mathrm{A} \omega\) (D) \(\mathrm{A}^{2} \omega \sin (\omega \mathrm{t} / 2)\)

Find a unit vector perpendicular to both \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) (A) $\left[\left(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}\right) / \mathrm{AB}\right]$ (B) $\left[\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{-}\right) /(\mathrm{AB} \sin \theta)\right]$ (C) $\left[\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right) /(\mathrm{AB} \cos \theta)\right]$ (D) $\left[\left(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}\right) /(\mathrm{AB} \sin \theta)\right]$

Equation of a projectile is given by \(\mathrm{y}=\mathrm{Ax}-\mathrm{Bx}^{2}\). Find the range for the particle. (A) \((\mathrm{A} / \mathrm{B})\) (B) \((\mathrm{A} / 4 \mathrm{~B})\) (C) \((\mathrm{A} / 2 \mathrm{~B})\) (D) \((2 \mathrm{~A} / \mathrm{B})\)

Which from the following is true? (A) $\cos \theta=\left[\left(\left|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right|\right) / \mathrm{AB}\right]$ (B) $\sin \theta=\left[\left(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}\right) / \mathrm{AB}\right]$ (C) $\tan \theta=\left[\left(\left|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right|\right) /\left(\mathrm{A}^{-}-\mathrm{B}^{-}\right)\right]$ (D) $\cot \theta=\left[\mathrm{AB} /\left(\left|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right|\right)\right]$
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