An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s after firing the arrow and the average speed of the arrow was 40 mis, what was the distance separating the archer and the target? Use 340 m/ s for the speed of sound.

Now, the archer fires the arrow, that, in a time [tex]t_{arrow}[/tex] travels the distance d with a speed [tex] v_{arrow}[/tex] of 40 m/s and hits the target. We can see that the equation will be:

[tex]v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d[/tex]

Immediately after this, the arrow produces a muffled sound, which will travel the distance d at 340 m/s in a time [tex] t_{sound} [/tex]. Obtaining :

[tex]v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d[/tex].

Finally, the sound reaches the archer, exactly 1 second after he fired the bow, so:

[tex]t_{arrow} + t _{sound} = 1 s[/tex].

This equation allows us to write:

[tex] t _{sound} = 1 s - t_{arrow}[/tex].

Plugging this relationship in the distance equation for the sound:

[tex] 340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d[/tex].

Now, we can replace d from the first equation, and obtain:

[tex] 40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow})[/tex].

Now, we can just work a little bit:

[tex] 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s[/tex].

Now, we can just plug this value into the first equation:

[tex] 40 \frac{m}{s} * t_{arrow} = d [/tex]

[tex] 40 \frac{m}{s} * 340/380 s = 35,79 s = d [/tex]