For an object moving with uniform acceleration, travelling 50m in 5^th sec, 70m in 7^th sec.
a) Its initial velocity is 5 m/s
b) Its acceleration is 20 m/s²
c) Its travels 100 m in 9th sec
d) Its average velocity during 9th sec is 90 m/s

${\text{S}}_5=\text{u}+\text{a}(5-\frac{1}{2})=50\Rightarrow \text{u}+\frac{99}{2}=50$

 $$\begin{gathered} S_7=u+a(7-1/2) = 70 \\ \Rightarrow u + 13a/2 = 70 \end{gathered}$$          

From above equations a = 10ms⁻²

u + 9a/2 = 50

u + 9(10)² =50

u = 50 − 45 = 5ms⁻¹

Distance travel in the 9^th second

S₉ = u + a(9 − 1/2)

S₉ = 5 + 10(17/2)

      =90 m

Final velocity after 8 sec

v = u + at = 5 + 8×10 = 85

(means during the 9^th second)

Final velocity after 9 seconds

v = u + at

   = 5 + 10×9

   = 95 m/s

$\overrightarrow{v_{avg}}$ = (85 + 95)/2 = 90m/s