A proint moving in a straight line travels in its second, ` 16 m, 28 m and 52 m` respectively , prove that point is moving with constant acceleration. Aso find the total distance moving by particle in ` 10 seconds.

Explanation

Let (u) be the initial velocity of a point and (a) be its acceleraton. Distance travelled in the nth second by a particle moving in a straight line with a constant accleration si
` D_n = u = a/2 ( 2 n-1)`
So, ` 16 = u + a/2 92 xx 2 -1)` …(i)
` 28 = u = a/2 (2 xx 5 -1)` ....(ii)
` and ` 52 =u + a/2 ( 2 xx 11 -1)` ....(iii) `
On solcing (i0, (ii) and (iii) , we get
` a=4 ms^(-2) , u= 10 ms^(-1)`
Now , ` S= ut + 1/2 t^2 = 10 xx 10 + 1/2 xx 4 xx 10^2`
`100+ 200 = 300 m`.