Figure is given a plot of the cooordinate of the body against time. From `t_(1)` to `t_(2)` the graph is a parabola.

Options

A. The acceleration from `t_(1)` to `t_(3)` is `(2(x_(1)-x_(0)))/(t_(1)(t_(3)-t_(1)))`
B. The initial velocity is `(x_(1)-x_(0))/(t_(1))`
C. The maximum x coordinates is `x_(1)+((x_(1)-x_(0))(t_(3)-t_(1)))/(t_(1))`
D. The body is speeding up at all times.

Answer

ABCD

Explanation

`u=(x_(1)-x_(0))/(t_(1))`
`a=(u-(-u))/(t_(3)-t_(1)) =(2(x_(1)-x_(0)))/(t_(1)(t_(3)-t_(1)))`
at `x_(max) V=0`
`v^(2)-u^(2)-2as`
`0^(2) =((x_(1)-x_(0))/(t_(1)))^(2) 2xx(2(x_(1)-x_(0)))/(t_(1)(t_(3)-t_(1)))xx(x-x_(1))`
`((x_(1)-x_(0))/(4t_(1)))(t_(3)-t_(1))=x-x_(1)`
`x=x_(1)+((x_(1)-x_(0))(t_(3)-t_(1)))/(4t_(1))`