(A) Given line have positive intercept but negative slope. So its equation can be written as, v = -mx + v0 …(1) where m = tanθ =
v0
x0
By differentiating with respect to time we get,
dv
dt
=−m
dx
dt
= -mv Now substituting the value of ‘v’ from equation (1) we get
dv
dx
= -m[ -mx + v0 ] = m2x−mv0 a = m2x−mv0 i.e., the graph between a and x should have positive slope but negative intercept on acceleration axis. Hence option (A) is correct.