A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the 5th second to that coverd in 5 seconds is

Question

A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the 5th second to that coverd in 5 seconds is (A)  (9/25) (B)  (3/25) (C)  (25/9) (D)  (1/25)

Tardigrade Admin · Accepted Answer

Distance covered in 5th second is D5 = 0+(a/2)(2×5-1) = (9a/2) Distance covered in 5 seconds is S5 = 0+(1/2) × a×52 = (25a/2) ∴ (D5/S5) = (9/25)

Q. A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the $5^{t h}$ second to that coverd in 5 seconds is

$\frac{9}{25}$

$\frac{3}{25}$

$\frac{25}{9}$

$\frac{1}{25}$

Distance covered in $5^{t h}$ second is
$D_{5} = 0 + \frac{a}{2} (2 \times 5 - 1) = \frac{9 a}{2}$
Distance covered in 5 seconds is
$S_{5} = 0 + \frac{1}{2} \times a \times 5^{2} = \frac{25 a}{2}$
$∴ \frac{D _{5}}{S _{5}} = \frac{9}{25}$

Q. A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the 5th second to that coverd in 5 seconds is

259​

253​

925​

251​