Question

the distance travelled by a body falling from rest in the first second and third second are in the ratio?​

Answer 1

$\boxed{\sf {\green{Answer}}}$

ratio = 1:5

$\boxed{\sf {\green{Explanation}}}$

Given:

• u = 0
• a = -g

To Find:

ratio of distance travelled by the body in 1st and 3rd second

Solution

w.k.t,

$\boxed{\bold{\pink{S_n = u + a(n-\dfrac{1}{2})}}}$

$\mathsf{S_1 = 0+(-10) (1-\dfrac{1}{2})}$

$\mathsf{S_1 = (-10) (\dfrac{1}{2})}$

$\mathsf{S_1 = -5m}$

Now,

$\mathsf{S_3 = 0+(-10) (3-\dfrac{1}{2})}$

$\mathsf{S_3 = (-10) (\dfrac{5}{2})}$

$\mathsf{S_3 = -25m}$

Ratio of distances

$\mathsf{\dfrac{S_1}{S_3} =\dfrac{-5m} {-25m}}$

$\mathsf{\dfrac{S_1}{S_3} =\dfrac{1} {5}}$

$\mathsf{S_1 : S_3=1 : 5 }$

Note:

-ve sign of displacement indicates that, Displacement is in downward direction

$\texttt{\blue{Aravind}\: \red{Reddy}....!}$

Answer 2

$\bf{\huge{\boxed{\underline{\mathbb{\green{Answer:}}}}}}$

GIVEN →

• u = 0
• u = 0a = -g

To Find →

ratio of distance travelled by the body in 1st and 3rd second

Solution →

BY PUTTING THE VALUES IN

THE FORMULA , NOW LETS SOLVE THE QUESTION ..........

$\boxed{\bold{\red{S_n = u + a(n-\dfrac{1}{2})}}}$

$\tt{S_1 = 0+(-10) (1-\dfrac{1}{2})}$

$\tt{S_1 = (-10) (\dfrac{1}{2})}$

$\tt{S_1 = -5m}$

Now,

$\tt{S_3 = 0+(-10) (3-\dfrac{1}{2})}$

$\tt{S_3 = (-10) (\dfrac{5}{2})}$

$\tt{S_3 = -25m}$

Ratio of distances

$\tt{\dfrac{S_1}{S_3} =\dfrac{-5m} {-25m}}$

$\tt{\dfrac{S_1}{S_3} =\dfrac{1} {5}}$

$\tt{S_1 : S_3=1 : 5 }$

Note:

-ve (negative) sign of displacement shows that, Displacement is in downward direction