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Q 3.     Abdul, while driving to school, computes the average speed for his trip to be 20\ km h^{-1}. On his return trip along the same route, there is less traffic and the average speed is 30\ km h^{-1}. What is the average speed for Abdul’s trip?

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Given, Average speed while going to school, v_1 = 20\ km h^{-1}

And Average speed while returning back from school, v_2 = 30\ km h^{-1}

Let the distance between starting point and school be x\ m

And time taken by Abdul during the two trips be t_1\ s\ and\ t_2\ s

We know, Speed = \frac{Distance}{Time}

\therefore v_1 = \frac{x}{t_1} = 20

And, \therefore v_2 = \frac{x}{t_2} = 30             -(i)

Now, Total distance that Abdul covers = x +x = 2x

And total time Abdul takes to cover this distance =t_1 + t_2

\\ \therefore v_{avg} = \frac{2x}{t_1+t_2} \\ = \frac{2x}{\frac{x}{20}+\frac{x}{30}} \\ \\ = \frac{60\times2x}{5x} \\ \\ = 24\ ms^{-1}

Therefore, the average speed for Abdul's trip is 24\ ms^{-1}

(Note: \frac{20+30}{2} =25\ ms^{-1} \neq 24\ ms^{-1})

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