The table shows the distance covered in successive seconds by a body accelerated uniformly from rest `{:("Time interval"(s) ,1st , 2nd, 3rd, 4th),("Distance"(cm), 2 ,6 ,10, 14):}`
The speed of the body at the end of 4th second (in `cms^(-1)` ) is

Options

A. 16
B. 8
C. 14
D. 4

Answer

A

Explanation

`x=1/2 at^2 [ :. a " is constant and " u=0]`
`t=1s, x=2cm :. 2=1/2 a(t)^2 implies a=4 cms^(-2), v=u+at implies (v)_(t=4s)=0+(4)(4)=16cms^(-1)`