Moving with uniform acceleration, a body covers 150m during  10s so that it covers 24m during  10th sec. Find the initial velocity

Let initial velocity of the body be u and ‘a’ be its acceleration
Given that  S=150m

t=10S

$\therefore$ From equation  $S=ut+\frac{1}{2}a{t}^2$
$150=u\left(10\right)+\frac{1}{2}a{\left(10\right)}^2$ 
$\Rightarrow u+5a=\mathrm{15..........}\left(1\right)$ 
And also Distance travelled in 10th sec is 24m i.e $n=10\&S_n=24m$

From equation 

$\begin{array}{l}\therefore S_n=u+a\left(n-\frac{1}{2}\right)\\ \Rightarrow 24=u+a\left(10-\frac{1}{2}\right)\\ \Rightarrow 24=u+\frac{19}{2}a\\ \Rightarrow 2u+19a=48\dots \dots \dots \dots \dots .\left(2\right)\end{array}$

On solving (1) & (2)
acceleration  $a=2m/s^2\mathrm{......}\left(3\right)$
And initial velocity  $u=5m/s$