The velocity-displacement graph of a particle moving along a straight line is shown. The most suitable acceleration-displacement graph will be

Options

A.
B.
C.
D.

Answer

A

Explanation

The velocity-displacement graph is given. Corresponding acceleration-displacement graph is required.
From v-x graph, the corresponding equation is
`v=(-v_0/x_0)x+v_0` (Pattern , y=mx+c)
`therefore (dv)/(dt)=(-v_0/x_0) (dx)/(dt)` or `a=(-v_0/x_0)v`
`a=(-v_0/x_0)xx[(-v_0/x_0)x+v_0]`
`a=(v_0/x_0)^2 x- v_0^2/x_0` (Pattern , y=mx+c)
Slope of the line = `(v_0/x_0)^2`
Slope is positive . Therefore `theta` is acute .
Intercept `=(-v_0^2)/x_0` . It is negative
The slope and intercept reveal that the graph (a) is the most suitable representation.