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 equation for the given ` v-x` graph is
` v = - (v_(0))/(x_(0)) + v_(0)` ….. (i)
`(dv)/(dx) = - ( v^(0))/(x_(0))`
`:. v(dv)/(dx) = - (v)/(x_(0)) xx v = -(v_(0))/(x_(0))[-(v_(0))/(x_(0))+ v_(0)] ` from (1)
`:. a = (v_(0)^(2))/(x_(0)^(2))x - (v_(0)^(2))/(x_(0))` ....(ii) `[ because a = v(dv)/(dx)]`
On comparing the equation (ii) with equation of a straight line
` y = mx + c`
We get`m = (v_(0)^(2))/(x_(0)^(2)) = +ve ` ,
i.e. ` tan theta = + ve` , i.e., `theta ` is acute.
Also `c = -(v_(0)^(2))/(x_(0)^(2))` ,
i.e., the `y - intercept ` is negative
The above conditions are satisfied in graph `(a)`.