When the distance travelled by a body is directly proportional to the square of time?
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According to the 2nd equation of motion,
s = ut + ½ at^2
Now, consider the initial velocity to be zero.
So, s = ½ at^2
Again, consider that the body is experiencing a vertical motion i.e. a free fall. (For example, the motion of a ball if it is dropped from the top of a high building.)
In this case, acceleration (a) = g (acceleration due to gravity, which is a constant)
g = 9.8 m/s^2, everywhere on the earth
So, s = ½ gt^2
(Let sign of proportionality = @)
So, s @ t^2 (as ½ and g are constants and get eliminated as proportionality sign is introduced)
So, for distance travelled by a body is directly proportional to time if :-
1.) The initial velocity is zero.
2.) The body experiences a free fall such that a = g.
s = ut + ½ at^2
Now, consider the initial velocity to be zero.
So, s = ½ at^2
Again, consider that the body is experiencing a vertical motion i.e. a free fall. (For example, the motion of a ball if it is dropped from the top of a high building.)
In this case, acceleration (a) = g (acceleration due to gravity, which is a constant)
g = 9.8 m/s^2, everywhere on the earth
So, s = ½ gt^2
(Let sign of proportionality = @)
So, s @ t^2 (as ½ and g are constants and get eliminated as proportionality sign is introduced)
So, for distance travelled by a body is directly proportional to time if :-
1.) The initial velocity is zero.
2.) The body experiences a free fall such that a = g.
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