The position of a particle along X-axis at time t is given by `x=2+t-3t^(2)`. The displacement and the distance travelled in the interval, t = 0 to t = 1 are respectively

Options

A. 2, 2
B. `-2, 2.5`
C. 0, 2
D. `-2, 2.1`

Answer

D

Explanation

`x=2+t-3t^(2), upsilon=(dx)/(dt)=1-6t`, velocity will become zero at time to `0=1-6 t_(0)` or `t_(0)=(1)/(6)s`,
Since, the given time t = 1s is greater than `t_(0)=(1)/(6)s`,
distance gt |displacement|
Displacement, `S=x_(f)-x_(i)`
`=(2+1-3)-(2+0-0)=-2m`
Distance, `d=|S_(0-t_(0))|+|S_(t-t_(0))|`
`=(u^(2))/(2|a|)+(1)/(2)|a|(t-t_(0))^(2)`
On comparing `upsilon=1-6t` with `upsilon=u+at`, we have
`u=1ms^(-1)` and `a=-6 ms^(-2)`
`therefore` Distance `=((1)^(2))/(2xx6)+(1)/(2)xx6xx(1-(1)/(6))^(2)=2.1m`