Question

A man 'X" is standing inside a lift which is starting from rest moving up with an acceleration g/2. After 2 seconds 'X' drops a particle, acceleration of particle as observed by 'X' will be (in m/s') g = 10 m's?

Answer 1

Answer:

Lift is accelerating downwards, so acceleration of lift w.r.t. ground  a  

lg

​  

=a

Acceleration of ball w.r.t. ground  a  

bg

​  

=g

Acceleration of ball w.r.t. lift   a  

bl

​  

=a  

bg

​  

−a  

lg

​  

=g−a

Explanation:

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Answer 2

The acceleration of the particle is 15m/s² downwards.

Given:-

Acceleration of lift = g/2

Time = 2sec

To Find:-

The acceleration of the particle.

Solution:-

We can simply calculate the acceleration of the particle by using the following process.

As

Acceleration of lift (a) = g/2

g = 10m/s²

Time (t) = 2sec

The acceleration of the particle (g') =?

Since lift and particle both are accelerating so we can find the acceleration of particle by using the relative acceleration approach.

acceleration of particle = g - acceleration of the lift

g' = - g - \frac{g}{2}

the lift$g' = - g - \frac{g}{2}$$g' = \frac{ - 2g - g}{2}$

the lift$g' = - g - \frac{g}{2}$$g' = \frac{ - 2g - g}{2}$$g' = \frac{ - 3g}{2}$

the lift$g' = - g - \frac{g}{2}$$g' = \frac{ - 2g - g}{2}$$g' = \frac{ - 3g}{2}$$g' = \frac{ - 3 \times 10}{2}$

the lift$g' = - g - \frac{g}{2}$$g' = \frac{ - 2g - g}{2}$$g' = \frac{ - 3g}{2}$$g' = \frac{ - 3 \times 10}{2}$$g' = \frac{ - 30}{2}$

the lift$g' = - g - \frac{g}{2}$$g' = \frac{ - 2g - g}{2}$$g' = \frac{ - 3g}{2}$$g' = \frac{ - 3 \times 10}{2}$$g' = \frac{ - 30}{2}$$g' = - 15$

the lift$g' = - g - \frac{g}{2}$$g' = \frac{ - 2g - g}{2}$$g' = \frac{ - 3g}{2}$$g' = \frac{ - 3 \times 10}{2}$$g' = \frac{ - 30}{2}$$g' = - 15$Hence, The acceleration of the particle is 15m/s² downwards.

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