Question:

A particle covers 3/4 th of total distance with speed v1 and next 1/4 th with speed v2. Find the average speed of the particle.

Solution:

$$Average \ speed = \frac{Total \ distance}{Total \ time \ taken}$$

Total distance is $${\frac{3}{4}(d)+\frac {1}{4}(d) =d}$$

Let the speed be v₁ for the 3/4 th distance and v₂ for the next 1/4 th distance.

Time taken to cover 3/4 th distance with velocity v₁ is $${t_1=\frac{3d}{4v_1}}$$

Time taken to cover next 1/4 th distance with velocity v₂ is $${t_2 =\frac {d}{4v_2}}$$

Hence average speed is given by $${\frac{4v_1v_2}{v_1+3v_2}}$$

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