Answer:
[tex]w=694.575\ m[/tex]
Explanation:
Given:
velocity of the sound, [tex]v=343\ m.s^{-1}[/tex]
time lag in echo form one wall of the valley, [tex]t= 1.55\ s[/tex]
time lag in echo form the other wall of the valley, [tex]t'=2.5\ s[/tex]
distance travelled by the sound in the first case:
[tex]d=v.t[/tex]
[tex]d=343\times 1.55[/tex]
[tex]d=531.65\ m[/tex]
Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.
[tex]s=\frac{d}{2}[/tex]
[tex]s=\frac{531.65}{2}[/tex]
[tex]s=265.825\ m[/tex]
distance travelled by the sound in the second case:
[tex]d'=v.t'[/tex]
[tex]d'=343\times 2.5[/tex]
[tex]d'=857.5\ m[/tex]
Since this is the distance covered by the sound while going form the source to the walls and then coming back from the wall to the source so the distance of the wall form the source will be half of the distance obtained above.
[tex]s'=\frac{d'}{2}[/tex]
[tex]s'=\frac{857.5}{2}[/tex]
[tex]s'=428.75\ m[/tex]
Now the width of valley:
[tex]w=s+s'[/tex]
[tex]w=265.825+428.75[/tex]
[tex]w=694.575\ m[/tex]