a rope stretched between two boats at rest .a sailor in first board pull the rope with the constant force of 100 Newton first boat with the sailor has a mass of 250 kg where as the mass of the second boat is double the mass of first one and if the initial distance between the boat is 30 then find the time taken for two boat to meet each other(neglect water resistance between boats and water)
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5
Given, force acts on both boat = 100
So, 250 * a1 = 100
a1 = 100/250
= 0.4
Given, mass of second boat is double.(i.e 250 * 2 = 500)
So, 500 * a1 = 100
a1 = 100/500
= 0.2
Relative acceleration = 0.4 + 0.2
= 0.6
We know that S = ut + 1/2 at^2
30 = 0 + 1/2(0.6)* t^2
30 = (0.3) * t^2
t^2 = 30/0.3
t^2 = 300/3
t^2 = 100
t = 10 seconds.
Hope this helps!
So, 250 * a1 = 100
a1 = 100/250
= 0.4
Given, mass of second boat is double.(i.e 250 * 2 = 500)
So, 500 * a1 = 100
a1 = 100/500
= 0.2
Relative acceleration = 0.4 + 0.2
= 0.6
We know that S = ut + 1/2 at^2
30 = 0 + 1/2(0.6)* t^2
30 = (0.3) * t^2
t^2 = 30/0.3
t^2 = 300/3
t^2 = 100
t = 10 seconds.
Hope this helps!
sanidhyasingla:
thank you very much
but what is relative acceleration
Relative acceleration formula is a = a1 - (- a2) = 0.4 + 0.2 = 0.6
Answered by
23
Answer:
Given, force acts on both boat = 100
So, 250 * a1 = 100
a1 = 100/250
= 0.4
Given, mass of second boat is double.(i.e 250 * 2 = 500)
So, 500 * a1 = 100
a1 = 100/500
= 0.2
Relative acceleration = 0.4 + 0.2
= 0.6
We know that S = ut + 1/2 at^2
30 = 0 + 1/2(0.6)* t^2
30 = (0.3) * t^2
t^2 = 30/0.3
t^2 = 300/3
t^2 = 100
t = 10 seconds.
Hope this helps!
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