Question

a body starts moving with constant acceleration then the ratio of distance travelled by the particle in fifth and sixth second is?

Answer 1

Answer:

Distance travelled in nth second  $s_n = u + \dfrac{a}{2}(2n-1)$

u = initial speed, a = acceleration, n = nth second

$s_5 = u + \dfrac{a}{2}(2 \times 5 -1) = u + \dfrac{a}{2}(9)\\s_6 = u + \dfrac{a}{2}(2 \times 6 -1) = u + \dfrac{a}{2}(11)\\\\ \dfrac{s_5}{s_6} = \dfrac{2u+9a}{2u+11a}$

Now, if u = 0, then

$\dfrac{s_5}{s_6} = \dfrac{9}{11}$