A particle starts from rest on a straight path, Its acceleration is linearly varying with time, such that velocity of the particle at t=2 sec and t=4 sec is `6ms^(-1)` respectively. Find acceleration of the particle at t=i sec.

Options

A. `1.5ms^(-2)`
B. `2 ms^(-2)`
C. `3 ms^(-2)`
D. `2.5ms^(-2)`

Explanation

Let `alphat+beta`
`v=(alphat^(3))/(2)+betat+gamma`
at `t=0 v=0rArrgamma=0`
`rArrv=(alphat^(2))/(2)+betat`
at t=2 v=6
`62alpha+2beta`
`alpha+beta=3`
at t=4 v=20
`20=8alpha+4beta`
`2alpha+beta=5`
from (i) `&` (ii)
`alpha=2`
`beta=1`
`therefore "at" t=1 a=2t+1=3`