EDIT
Sorry but this development contains an error (replacement of bound2 by bound2a). Thanks to bbgodfrey for notifying me.
It turns out that the original bound2 is ill defined in the OP as it contains a derivative of the same order as the differential equation. Only if one makes deliberately the same error as I did (writing dx/dt instaed of dx/ds on the right hand side of bound2), declaring it a modification, a reasonable non trivial result appears.
Original text
In the first step we write down the mathematical formulation of the problem. Then we transform the equations to Mathematica, and solve the problem.
Mathematical formulation
The function to be determined is $x(s,t)$. It is governed by this nonlinear partial differential equation
$$\text{eq}=\frac{\partial ^2x(s,t)}{\partial t^2}\text{==}\frac{\partial ^2x(s,t)}{\partial s^2}/\left(1+ \frac{\partial x(s,t)}{\partial s}\right)^{7/2}$$
The initial conditions at $t=0$ are
$$\text{init1}= x(s,t)\text{/.}t\to 0==0$$
$$\text{init2} = \frac{\partial x(s,t)}{\partial t}\text{/.}t\to 0==0$$
The boundary conditions at specified values of s are
$$\text{bound1}= x(s,t)\text{/.}s\to 0==0$$
$$\text{bound2}=\frac{\partial ^2x(s,t)}{\partial t^2}\text{/.}s\to 1==-1 + 5\left/\left(1+ \frac{\partial x(s,t)}{\partial s}\text{/.}s\to 1\right)^{5/2}\right.$$
The condition bound2 can be simplified. In fact, defining the function
$$\text{bound2a}=\ g(t)=\frac{\partial x(s,t)}{\partial t}\text{/.}s\to 1$$
which at the same time is the simplified boundary condition, bound2 reads an an ODE for $g$
$$\text{eqg}=\frac{\ d g(t)}{\ d t}==-1 + 5\left/\left(1+ \ g(t) \right)^{5/2}\right.$$
The initial condition for $g$ is taken from init2 to be
$$\text{initg}=g(0) == 0$$
Formulation in Mathematica
In order to have the boundary conditions complete we will solve first the ODE eqg.
It is convenient to do this without imposing an initial condition. This will be done subsequently numerically.
Here we go
solg = DSolve[g'[t] == -1 + 5/(1 + g[t])^(5/2), g[t], t]
(*
Out[150]= {{g[t] -> InverseFunction[
1/10 (-2 5^(2/5) Sqrt[10 - 2 Sqrt[5]]
ArcTan[(5 - 5 Sqrt[5] + 4 5^(4/5) Sqrt[1 + #1])/(
5 Sqrt[2 (5 + Sqrt[5])])] +
2 5^(2/5) Sqrt[2 (5 + Sqrt[5])]
ArcTan[(5 + 5 Sqrt[5] + 4 5^(4/5) Sqrt[1 + #1])/(
5 Sqrt[10 - 2 Sqrt[5]])] +
4 5^(2/5) Log[5 - 5^(4/5) Sqrt[1 + #1]] -
5^(2/5) (1 + Sqrt[5]) Log[
5 - 1/2 5^(4/5) (-1 + Sqrt[5]) Sqrt[1 + #1] + 5^(3/5) (1 + #1)] +
5^(2/5) (-1 + Sqrt[5]) Log[
5 + 1/2 5^(4/5) (1 + Sqrt[5]) Sqrt[1 + #1] + 5^(3/5) (1 + #1)] +
10 #1) &][-t + C[1]]}}
*)
In order to comply with initg
we have to adjust the constant C[1]
properly. This is done numerically thus
u0 = u /. FindRoot[0 == ((g[t] /. solg) /. t -> 0 /. C[1] -> u), {u, 1}]
(*
Out[213]= 1.19413
*)
The function is therefore (called gg henceforth)
gg[t_] := (g[t] /. solg /. C[1] -> u0)
Plot[gg[t], {t, -1, 5}, PlotRange -> {-1, 1},
PlotLabel -> "The function gg(t)"]
(* 150602_plot _gg.jpg *)

Now we can write down the complete equations and conditions
eq = D[x[s, t], {t, 2}] == D[x[s, t], {s, 2}] 1/(1 + D[x[s, t], s])^(7/2);
init1 = x[s, 0] == 0;
init2 = (D[x[s, t], t] /. t -> 0) == 0;
bound1 = x[0, t] == 0;
bound2a = (D[x[s, t], t] /. s -> 1) == gg [t];
And find the numerical solution
sol = NDSolve[{eq, init1, init2, bound1, bound2a}, {x[s, t]}, {s, 0, 1}, {t,
0, 1}]
(*
Out[224]= {{x[s, t] -> \!\(\*
TagBox[
RowBox[{"InterpolatingFunction", "[",
RowBox[{
RowBox[{"{",
RowBox[{
RowBox[{"{",
RowBox[{"0.`", ",", "1.`"}], "}"}], ",",
RowBox[{"{",
RowBox[{"0.`", ",", "1.`"}], "}"}]}], "}"}], ",", "\<\"<>\"\>"}], "]"}],
False,
Editable->False]\)[s, t]}}
*)
Here's the graph of x(s,t)
Plot3D[Evaluate[x[s, t] /. sol], {t, 0, 1}, {s, 0, 1}, PlotRange -> All,
PlotLabel -> "The solution x(s,t)", AxesLabel -> {"t", "s", "x(s,t)"}]
(* 150602_Plot3D _xst.jpg *)

Conclusion
The problem can be solved in Mathematica once we analyse carefully the equation and initial and boundary conditions (which is not trivial in this case).
Despite the first impression based on a linearization the solution is not trivially zero. On the other hand is looks rather innocent and there is not much "suspense" in it.
NDSolve::bdord: "Boundary condition 1 + Derivative[0, 2][x][1, t] - 5/(1 + Derivative[1, 0][x][1, t])^(5/2) should have derivatives of order lower than the differential order of the partial differential equation"
$\endgroup$