# 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?

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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