Answer:
Total distance covered equals [tex]\frac{48}{7}feet[/tex]
Explanation:
The situation is represented in the attached figure
Distance in first bounce = [tex]d_{1}=2\times 3ft[/tex]
Distance in second bounce =[tex]d_{2}=2\times \frac{3}{8}ft[/tex]
Distance in third bounce=[tex]d_{3}=2\times \frac{3}{8^{2}}ft[/tex]
Thus the total distance covered = [tex]d_{1}+d_{2}+d_{3}+...[/tex]
Applying values we get
Total distance covered = [tex]2\times 3+2\times \frac{3}{8}+2\times \frac{3}{8^{2}}+2\times \frac{3}{8^{3}}+....\\\\=6(1+\frac{1}{8}+\frac{1}{8^{2}}+\frac{1}{8^{3}}+...)[/tex]
Summing the infinite geometric series we get total distance covered as[tex]S_{\infty }=\frac{a}{1-r}[/tex]
[tex]D=6(\frac{1}{1-\frac{1}{8}})\\\\D=\frac{48}{7}feet[/tex]
