Question

a body start from rest. what is the ratio of distance travelled by the body during fourth and third second

Answer 1

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Answer 2

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

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

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

$\green{\underline \bold{Given :}}$

$\tt: \implies Initial \: velocity = 0 \: m/s$

$\red{\underline \bold{To \: Find :}}$

$\tt: \implies \frac{ s_{4} }{ s_{3} } =?$

• Given Question

$\bold{As \: we \: know \: that}$

$\tt: \implies s_{n} = u + \frac{a}{2} (2n - 1)$

$\tt: \implies s_{4} = 0 + \frac{a}{2} (2 \times 4 - 1)$

$\tt: \implies s_{4} = \frac{a}{2} (8 - 1)$

$\tt: \implies s_{4} = \frac{7a}{2} - - - - - (1)$

$\bold{As \: we \: know \: that}$

$\tt: \implies s_{n} = u + \frac{a}{2} (2n - 1)$

$\tt: \implies s_{3} = 0 + \frac{a}{2} (2 \times 3 - 1)$

$\tt: \implies s_{3} = \frac{a}{2} (6 - 1)$

$\tt: \implies s_{3} = \frac{5a}{2} - - - - - (2)$

$\bold{For \: Ratio : }$

$\tt: \implies \frac{ s_{4} }{ s_{3} } = \frac{ \frac{7a}{2} }{ \frac{5a}{2} }$

$\tt: \implies \frac{ s_{4} }{ s_{3} } = \frac{7a}{2} \times \frac{2}{5a}$

$\tt: \implies \frac{ s_{4} }{ s_{3} } = \frac{7}{5}$

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