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Answer:
Answer:
Given :-
A body moving with rest of uniform acceleration travels a distance 'X' in the first t seconds and travels the distance 'Y' with same acceleration in next 2t seconds.
To Find :-
What is the relationship.
Formula Used :
\clubsuit♣ Second Equation Of Motion :
\begin{gathered}\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2}at^2}}}\\\end{gathered}↦s=ut+21at2
where,
s = Distance Covered
u = Initial Velocity
t = Time
a = Acceleration
Solution :-
\begin{gathered}{\normalsize{\bold{\purple{\underline{\bigstar\: In\: the\: first\: case\: :-}}}}}\\\end{gathered}★Inthefirstcase:−
Given :
Distance Covered = X
Initial Velocity = 0 m/s
Time = t seconds
According to the question by using the formula we get,
\implies \sf X =\: 0(t) + \dfrac{1}{2} \times at^2⟹X=0(t)+21×at2
\implies \sf X =\: 0 + \dfrac{1}{2} \times at^2⟹X=0+21×at2
\implies \sf X =\: \dfrac{1}{2} \times at^2⟹X=21×at2
\begin{gathered}\implies \sf\bold{\green{X =\: 0.5at^2\: ------\: (Equation\: No\: 1)}}\\\end{gathered}⟹X=0.5at2−−−−−−(EquationNo1)
\begin{gathered}{\normalsize{\bold{\purple{\underline{\bigstar\: In\: the\: second\: case\: :-}}}}}\\\end{gathered}★Inthesecondcase:−
Given :
Distance Covered = Y
Initial Velocity = 0 m/s
Time = 2t seconds
According to the question by using the formula we get,
\begin{gathered}\implies \sf Y =\: \bigg(0(t) + \dfrac{1}{2} \times a \times (2t)^2\bigg) - 0.5at^2\\\end{gathered}⟹Y=(0(t)+21×a×(2t)2)−0.5at2
\begin{gathered}\implies \sf Y =\: \bigg(0 + \dfrac{1}{2} \times a \times 2t \times 2t\bigg) - 0.5at^2\\\end{gathered}⟹Y=(0+21×a×2t×2t)−0.5at2
\begin{gathered}\implies \sf Y =\: \bigg(0 + \dfrac{1}{2} \times a \times 4t^2\bigg) - 0.5at^2\\\end{gathered}⟹Y=(0+21×a×4t2)−0.5at2
\begin{gathered}\implies \sf Y =\: \bigg(\dfrac{1}{\cancel{2}} \times {\cancel{4}}at^2\bigg) - 0.5at^2\\\end{gathered}⟹Y=(21×4at2)−0.5at2
\implies \sf Y =\: 2at^2 - 0.5at^2⟹Y=2at2−0.5at2
\implies \sf Y =\: 1.5at^2⟹Y=1.5at2
\begin{gathered}\implies \sf\bold{\green{Y =\: 1.5at^2\: ------\: (Equation\: No\: 2)}}\\\end{gathered}⟹Y=1.5at2−−−−−−(EquationNo2)
Now, by dividing the equation no 2 by equation no 1 we get,
\longrightarrow \sf \dfrac{Y}{X} =\: \dfrac{1.5\cancel{at^2}}{0.5\cancel{at^2}}⟶XY=0.5at21.5at2
\longrightarrow \sf \dfrac{Y}{X} =\: \dfrac{1.5}{0.5}⟶XY=0.51.5
\longrightarrow \sf \dfrac{Y}{X} =\: \dfrac{\dfrac{15}{10}}{\dfrac{5}{10}}⟶XY=105