JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Kinematics

The velocity-distance graph is a straight line of negative slope, $v=v_0-kx$ with $k>0$. Acceleration along the path is $a=v\frac{dv}{dx}=(v_0-kx)(-k)=k^{2}x-k{v}_0$. This is a straight line in $x$ with positive slope $k^2$ and a negative intercept $-k{v}_0$ at $x=0$, crossing zero at $x=v_0/k$. The acceleration-distance graph is the straight line starting below the axis and rising with positive slope.