Question

A body starts from rest what is what is the ratio of the distance travelled by the body during fourth and third second

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{s_{4}:s_{3}=7:5}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Initial \: velocity = 0 \: m/s \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies \frac{ s_{4} }{ s_{3} } =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = u + \frac{a}{2} (2n - 1) \\ \\ \tt: \implies s_{4} = 0 + \frac{a}{2} (2 \times 4 - 1) \\ \\ \tt: \implies s_{4} = \frac{a}{2} (8 - 1) \\ \\ \tt: \implies s_{4} = \frac{7a}{2} - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = u + \frac{a}{2} (2n - 1) \\ \\ \tt: \implies s_{3} = 0 + \frac{a}{2} (2 \times 3 - 1) \\ \\ \tt: \implies s_{3} = \frac{a}{2} (6 - 1) \\ \\ \tt: \implies s_{3} = \frac{5a}{2} - - - - - (2) \\ \\ \bold{For \: Ratio : } \\ \tt: \implies \frac{ s_{4} }{ s_{3} } = \frac{ \frac{7a}{2} }{ \frac{5a}{2} } \\ \\ \tt: \implies \frac{ s_{4} }{ s_{3} } = \frac{7a}{2} \times \frac{2}{5a} \\ \\ \tt: \implies \frac{ s_{4} }{ s_{3} } = \frac{7}{5} \\ \\ \green{\tt: \implies { s_{4} } : { s_{3} } = 7 : 5}$