We study the Cauchy problem \[ { u t = ϕ ( u x ) x , ( x , t ) ∈ R × R + , u ( ⋅ , 0 ) = f \left \{ \begin {gathered} {u_t} = \phi {({u_x})_x},\qquad (x,t) \in {\mathbf {R}} \times {{\mathbf {R}}_ + }, \hfill \\ u( \cdot ,0) = f \hfill \\ \end {gathered} \right . \] with the piecewise linear constitutive function ϕ ( ξ ) = ξ + = max ( 0 , ξ ) \phi (\xi ) = {\xi _ + } = \max (0,\xi ) and with smooth initial data f f which satisfy x f ′ ( x ) ⩾ 0 xf’(x) \geqslant 0 , x ∈ R x \in {\mathbf {R}} , and f ( 0 ) > 0 f(0) > 0 . We prove that free boundary s s , given by u x ( s ( t ) + , t ) = 0 {u_x}(s{(t)^ + },t) = 0 , is of the form \[ s ( t ) = − κ t + o ( t ) , t → 0 + , s(t) = - \kappa \sqrt t + o\left ( {\sqrt t } \right ),\qquad t \to {0^ + }, \] where the constant κ = 0.9034 … \kappa = 0.9034 \ldots is the (numerical) solution of a particular nonlinear equation. Moreover, we show that for any α ∈ ( 0 , 1 / 2 ) \alpha \in (0,1/2) , \[ | d 2 d t 2 f ( s ( t ) ) | = O ( t α − 1 ) , t → 0 + . \left | {\frac {{{d^2}}} {{d{t^2}}}f(s(t))} \right | = O({t^{\alpha - 1}}),\qquad t \to {0^ + }. \] The proof involves the analysis of a nonlinear singular integral equation.