A diving bell is lowered into the sea at a speed of 6m/s and comes to rest with uniform retardation at a distance of 20m below the surface. What is it's retardation?
What you are calling retardation is usually called deceleration. Let its value be a.
Vo = sqrt(2 a X)
X = 20 m
Vo = initial velocity = 6 m/s
6 = sqrt(2*a*20)
36 = 40 a
a = 0.9 m/s^2
To find the retardation of the diving bell, we can use the kinematic equation that relates distance, initial velocity, final velocity, and acceleration (retardation in this case).
The equation we'll use is:
\[v^2 = u^2 + 2as\]
Where:
- \(v\) is the final velocity (which is zero in this case since the diving bell comes to rest)
- \(u\) is the initial velocity (given as 6 m/s)
- \(a\) is the acceleration (retardation)
- \(s\) is the distance traveled (given as 20 m)
Plugging in the given values into the equation, we can solve for \(a\):
\[0 = (6)^2 + 2a(20)\]
Simplifying the equation:
\[0 = 36 + 40a\]
Rearranging the equation to isolate the retardation:
\[40a = -36\]
\[a = \frac{-36}{40}\]
Simplifying:
\[a = -0.9 \, \text{m/s}^2\]
Therefore, the retardation of the diving bell is -0.9 m/s^2. The negative sign indicates that the bell is slowing down.