Question

A body starts from rest, what is the ratio of the
distance travelled by the body during the 4th
and 3rd seconds?
(a) 5 (b) 7
7 5

(c) 3 (d) 7
3 7

Answer 1

Answer:

Distance traveled by a free a free falling body in nth second =

distance traveled in n secs−distance traveled in n−1 secs.

$$=0*n+\dfrac { 1 }{ 2 } g{ n }^{ 2 }- 0*(n-1)+\dfrac { 1 }{ 2 } g{ (n-1) }^{ 2 }\\ =\dfrac { g }{ 2 } ({ n }^{ 2 } - ({ n }^{ 2 }+1-2n))\\ =\dfrac { g }{ 2 } (2n-1)$$

ratio of distance traveled in 4th sec. to 3rd sec $$= \dfrac { 2\times 4-1 }{ 2\times 3-1 } =\dfrac { 7 }{ 5 } $$

Answer 2

Answer:

A body starts from rest, what is the ratio of the

distance travelled by the body during the 4th

and 3rd seconds?

(a) 5 ✔️✔️

(b) 7

7 5

(c) 3

(d) 7

3 7

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