Answer :
Answer:
1:√2
Explanation:
S1 = S2 (Distance traversed by x = s1,by y = s2)
u = 0 for both
To find: time-x/time-y
a1 = 2a2
By 2nd eqn of motion
S1 = ut +1/2a1t1²= 1/2*2a2*t2²
S2 = ut + 1/2a2t2= 1/2*a2*t
S1 = S2
1/2*2a2*t1=1/2*a2*t2
2*t1²= t2²
t1² = t2²/2
t1²/t2² = 1/2
(t1/t2)² = 1/2
t1/t2 = √(1/2) = 1:√2
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
