Question

a car sta4t from rest qnd mives with constant acceleration the ratio of dis covered in 3 rd sec to that overed in 4 ses is​

Answer 1

Question:

A car starts from rest and moves with constant acceleration. The ratio of distance covered in 3rd second to that covered in 4th second is?

Answer:

$\bigstar{\bold{Ratio\:is\:9:16}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Initial Velocity = 0 m/s
• Body moving with constant acceleration, Δa = 0

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Ratio of distance covered in 3rd second to that of 4th second.

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ By the second equation of motion,

  s = ut + 1/2 × a × t²

→ Distance covered in 3rd second =

  s₃ = 0×3 + 1/2× a× 3²

  $s_3\:=\:\dfrac{9}{2} a$

→ Distance covered in 4th second =

  s₄ = 0×4 + 1/2× a × 4²

  s₄ = 8a

→ The ratio of distance covered in 3rd second to 4th second =

→ $\dfrac{s_3}{s_4}\:=\:\dfrac{\dfrac{9a}{2} }{8a}$

  $\dfrac{s_3}{s_4}\:=\:\dfrac{9 \cancel{a}}{2} \times \dfrac{1}{8 \cancel{a}}$

 $\dfrac{s_3}{s_4} \:=\:\dfrac{9}{16}$

$\boxed{\bold{Ratio\:is\:9:16}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The three equations of motion are:

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u² = 2as