Motion in a Straight Line

For the given velocity-displacement graph,Intercept, c=v0
Slope, m=−‌

v0

x0

∴ Slope, m=‌

dv

dx

=−‌

v0

x0

....(i)
For given diagram, the equation of line,
y=mx+c
v=−‌

v0

x0

x+v0 ...(ii)
Acceleration, a=‌

dv

dx

×‌

dx

dt

a=v‌

dv

dx

a=(−‌

v0

x0

x+v0)(−‌

v0

x0

) [From Eqs. (i) and (ii)]
⇒‌‌a=(‌

v0

x0

)2x−‌

v02

x0

Hence, the intercept is negative and the slope is positive, so the correct graph is given in option (c).