Answer:
t1 : t2 = 1 : √2
Explanation:
From equation of motion,
[tex]s = ut + \frac{1}{2}a {t}^{2} [/tex]
Since both starts from rest u = 0 and same distance s.
[tex]s = \frac{1}{2}a {t}^{2} [/tex]
Let
[tex] t_{1} \: be \: time \: taken \: by \: x \: and \: t_{2} \: be \: time \: taken \: by \: y[/tex]
[tex] s_{1} = \frac{1}{2} a_{1} t_{1} [/tex]
[tex] s_{2} = \frac{1}{2} a_{2}t_{2} [/tex]
Since
[tex] s_{1} = s_{2} \: and \: a_{1} = 2a_{2}[/tex]
[tex] \frac{1}{2} \times 2a_{2} { t_{1} }^{2} = \frac{1}{2} a_{2} { t_{2} }^{2} [/tex]
[tex]2 { t_{1} }^{2} = { t_{1} }^{2} [/tex]
[tex] \frac{ { t_{1} }^{2} }{ { t_{2} }^{2} } = \frac{1}{2} [/tex]
[tex] \frac{ t_{1} }{ t_{2} } = \frac{1}{ \sqrt{2} } [/tex]
t1 : t2 = 1 : √2
