A point initially at rest moves along x-axis. Its acceleration varies with time as `a = (6t + 5) m//s^(2)`. If it starts from origin, the distance covered in 2 s is:

Options

A. 20m
B. 18m
C. 16m
D. 25m

Answer

B

Explanation

Given, acceleration `a=6t+5`
`therefore a=(d upsilon)/(dt)=6t+5, d upsilon = (6t+5)dt`
`upsilon=3t^(2)+5t+C`, where C is a constant of integration.
When `t=0, upsilon = 0` so C = 0
`therefore upsilon = (d upsilon s)/(dt)=3t^(2)+5t` or `ds = (3t^(2)+5)dt`
Integrating it within the conditions of motion, i.r., as t changes from 0 to 2s, s changes from 0 to s, we have
`int_(0)^(s) ds=int_(0)^(2)(3t^(2)+5t)dt`
`therefore s=[t^(2)+(5)/(2)t^(2)]_(0)^(2)=8+10=18 m`