Two particles X and Y starting from rest cover the same distance. The acceleration of X is twice that of Y. the ratio of the time taken by X to that taken by Y is
A
\(\frac{1}{2}\)
B
\(\frac{2}{\sqrt2}\)
C
\(\frac{1}{\sqrt2}\)
D
\(\sqrt 2\)
E
4
Verified Correct Answer: C.
Physical Principles & Numerical Solution
Given \( a_X = 2a_Y \) and both start from rest covering the same distance \( s \):
For X:
\(s = a_Y t_X^2\)
For Y:
\(s = \frac{1}{2} a_Y t_Y^2\)
Setting them equal:
\(a_Y t_X^2 = \frac{1}{2} a_Y t_Y^2 \implies t_X^2 = \frac{1}{2} t_Y^2\)
Thus: \(\frac{t_X}{t_Y} = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad \frac{1}{\sqrt{2}}\)
Source: Myschool Past Questions Archive (Record #29021)