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Question

If a car covers $$\dfrac{2}{5^{th}}$$ of the total distance with $$v_{1}$$ speed and $$\dfrac{3}{5^{th}}$$ distance with $$v_{2}$$ then average speed:


A
 12v1v2
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B
v1+v22
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C
 2v1v2v1+v2
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D
5v1v23v1+2v2
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Solution

The correct option is D $$\dfrac{5v_{1}v_{2}}{3v_{1}+2v_{2}}$$
Time taken to cover $$\dfrac{2}{5}th$$ distance:$$t_1=\dfrac{2}{5}\dfrac{s}{v_1}=\dfrac{distance}{velocity}$$ where s is total distance
Time taken to cover $$\dfrac{3}{5}th$$ distance:$$t_1=\dfrac{3}{5}\dfrac{s}{v_2}$$
Average speed $$=\dfrac{total \quad distance}{total \quad time}$$
$$V_{mg}=\dfrac{s}{t_1+t_2}$$
$$=\dfrac{s}{\dfrac{2s}{5v_1}+\dfrac{3s}{5v_2}}$$
$$=\dfrac{5v_1v_2s}{s(2v_2s+3v_1s)}$$
$$\dfrac{5v_1v_2}{3v_1+2v_2}$$

Physics

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