A truck starts from rest and accelerate uniformly with `2 ms^(-2)`. At t = 10 s a stone is dropped by a person standing on the top of the truck (6 m high from ground). What are the (a) velocity and (b) acceleration of the stone at t = 11 s ? Neglect air resistance.

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as per the question.

### Given Data:
- Acceleration of the truck, \( a = 2 \, \text{m/s}^2 \)
- Initial velocity of the truck, \( u = 0 \, \text{m/s} \) (starts from rest)
- Time when the stone is dropped, \( t = 10 \, \text{s} \)
- Height of the truck from the ground, \( h = 6 \, \text{m} \)
- Time at which we need to find the stone's velocity and acceleration, \( t = 11 \, \text{s} \)
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