(a) The speed of a moving bus reduced from 45m/s to 5m/s with a uniform retardation of 10m/s\(^2\). Calculate the distance covered. (b) A bucket full of wat...

Answer Details

(a) The initial velocity of the bus is 45 m/s and the final velocity is 5 m/s, and the retardation is 10 m/s². We can use the equation of motion to find the distance covered. The equation is:

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance covered.

Substituting the given values, we get:

5² = 45² + 2(-10)s

25 = 2025 - 20s

20s = 2000

s = 100 meters

Therefore, the distance covered by the bus is 100 meters.

(b) The weight of the bucket is given by W = mg, where m is the mass of the bucket and g is the acceleration due to gravity. The tension in the rope is equal to the weight of the bucket, so we have:

T = W = mg

The net force acting on the bucket is given by F = T - mg, where F is the force accelerating the bucket.

Using Newton's second law of motion, F = ma, where a is the acceleration of the bucket. Substituting the values, we get:

ma = T - mg

a = (T - mg) / m

Substituting the given values, we get:

a = (240 - 16 x 10) / 16

a = 10 m/s²

Therefore, the acceleration of the bucket is 10 m/s².