Given that,
velocity, v = 7.8 ft/s
Radius, r = 1.25 ft
To find,
a. What is the acceleration of the bucket?
b. What is the velocity if the acceleration is 25 ft/sec²?
Solution,
(a) The centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{r}[/tex]
Put v = 7.8 ft/s and r = 1.25 ft
[tex]a=\dfrac{(7.8)^2}{1.25}\\\\=48.672\ ft/s^2[/tex]
(b) Put a = 25 ft/s² to find velocity.
[tex]v=\sqrt{ar} \\\\v=\sqrt{25\times 1.25} \\\\v=5.59\ ft/s[/tex]
Hence, this is the required solution.
