Question

The ratio of the distance cut in 5th and 3rd second to the speed starting at a constant acceleration of 8ms ^ -2 =​

Answer 1

$\color{darkblue}\underline{\underline{\sf Given-}}$

• Distance cut in 5th and 3rd second to the speed starting at a Constant Acceleration 8m/s²

$\color{darkblue}\underline{\underline{\sf To \: Find-}}$

• Ratio of Distance cut in 5th and 3rd seconds ${\sf \left(\dfrac{s_5}{s_3}\right)}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\color{green}\underline{\underline{\sf Formula \: Used-}}$

$\color{violet}\blacksquare\underline{\boxed{\sf s_n=u+\dfrac{a}{2}(2n-1)}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

➩ Distance cut in 5th second

$\implies{\sf s_5=0+\dfrac{8}{2}(2×5-1)}$

$\implies{\sf s_5=\dfrac{8}{2}×11}$

$\color{orange}\implies{\sf s_5=\dfrac{88}{2} }$

➩ Distance cut in 3rd second

$\implies{\sf s_3=0+\dfrac{8}{2}(2×3-1)}$

$\implies{\sf s_3=\dfrac{8}{2}×1 }$

$\color{orange}\implies{\sf s_3=\dfrac{8}{2}}$

$\underline{\sf Ratio \: of \: distances-}$

$\implies{\sf \dfrac{s_5}{s_3}=\dfrac{\dfrac{88}{2}}{\dfrac{8}{2}} }$

$\implies{\sf \dfrac{s_5}{s_3}=\dfrac{88}{8}}$

$\implies{\sf \dfrac{s_5}{s_3}=\dfrac{11}{1}}$

$\color{red}\implies{\sf s_5:s_3=11:1}$

$\color{darkblue}\underline{\underline{\sf Answer-}}$

Ratio of Distances is $\color{red}{\sf s_5:s_3=11:1}$