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A man starts from rest with an acceleration 1m/s^2 at t=0 .At t=3^1/2 it appears to him that rain falls with the velocity 3m/s vertically downwards. The velocity of actual rain falls is ?
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A man starts from rest with an acceleration 1m/s^2 at t=0 .At t=3^1/2 ...
Let the velocity of actual rain is 

A man starts from rest with an acceleration 1 m/s^2at t = 0.

velocity of man at t = 3sec ,  = 1/2 * at^2 i = 1/2 *1 * 9 = 4.5 i m/s

velocity of rain with respect to main, 

now, 

or, -3j = (a i + b j) - 4.5 i

or, - 3j = (a - 4.5)i + b j

comparing both sides,

a = 4.5 and b = -3.

hence, velocity of rain, 

and magnitude of  = √{4.5^2+ 3^2} = √{20.25 + 9} = √{29.25} = 5.4 m/s


Community Answer
A man starts from rest with an acceleration 1m/s^2 at t=0 .At t=3^1/2 ...
Problem Statement:
A man starts from rest with an acceleration of 1 m/s^2 at t=0. At t=3^1/2, it appears to him that rain falls with a velocity of 3 m/s vertically downwards. Determine the velocity of the actual rain falls.

Solution:

Given:
- Initial velocity of the man, u = 0 m/s
- Acceleration of the man, a = 1 m/s^2
- Time at which the man observes the rain, t = 3^1/2 s
- Apparent velocity of rain seen by the man, v = 3 m/s

Calculating the actual velocity of rain:
- Let the actual velocity of rain be V
- The apparent velocity of rain as seen by the man is the vector sum of the actual velocity of rain and the man's velocity at that instant
- v = V + u + at
- 3 = V + 0 + 1 * 3^1/2
- V = 3 - 3^1/2 m/s

Result:
The actual velocity of the rain falls is 3 - 3^1/2 m/s.
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A man starts from rest with an acceleration 1m/s^2 at t=0 .At t=3^1/2 it appears to him that rain falls with the velocity 3m/s vertically downwards. The velocity of actual rain falls is ?
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