A body starts from rest, what is the ratio of the distance travelled by the during the 4 textth and 3 textrd second ?

Question

A body starts from rest, what is the ratio of the distance travelled by the during the 4 textth and 3 textrd second ? (A) (7/5) (B) (5/7) (C) (7/3) (D) (3/7)

Tardigrade Admin · Accepted Answer

Distance covered in n textth second is given by Sn = u + (a/2) (2n - 1) Here, u = 0 ∴ S4 = 0 + (a/2) (2 × 4 - 1)=(7a/2) S3 = 0 + (a/2) (2 × 3 - 1)=(5a/2) ∴ (S4/S3)=(7/5)

Q. A body starts from rest, what is the ratio of the distance travelled by the during the $4^{th}$ and $3^{rd}$ second ?

$\frac{7}{5}$

$\frac{5}{7}$

$\frac{7}{3}$

$\frac{3}{7}$

Distance covered in $n^{th}$ second is given by
$S_{n} = u + \frac{a}{2} (2 n - 1)$
Here, u = 0
$∴ S_{4} = 0 + \frac{a}{2} (2 \times 4 - 1) = \frac{7 a}{2}$
$S_{3} = 0 + \frac{a}{2} (2 \times 3 - 1) = \frac{5 a}{2}$
$∴ \frac{S _{4}}{S _{3}} = \frac{7}{5}$

Q. A body starts from rest, what is the ratio of the distance travelled by the during the 4th and 3rd second ?

57​

75​

37​

73​