How to calculate the velocity of this body? 
Graph of a train (the body) is provided which starts from rest. What is the velocity after the train has 10 metres displacement?


Thats the only things provided for the question, please help me out here, I find that more variables will be required but the book states that no more variables are required. 
The answer key says that the answer is 10 m/s only! I am totally unable to understand how.
 A: We know that $v^2-u^2=2ax$ is valid for constant acceleration. So, what are we waiting for? Lets find how this apply this.
We will use $v^2-u^2=2ax$ for a $very$ small time(or distance) for which we will say acceleration is constant. 
$$v_0^2-u_0^2=2a_0x_0$$
$$v_1^2-u_1^2=2a_1x_1$$
$$v_2^2-u_2^2=2a_2x_2$$
$$\cdots$$
$$v_f^2-u_f^2=2a_fx_f$$
Note that $v_0=u_1$, $v_1=u_2$ as the intervals are directly after the time where previous equation was applied.
Add all of them. 
$$v_f^2-u_0^2=2(a_0x_0+...a_fx_f)$$
Now look at the graph. Each term in right hand side represents area of the very small part of graph :
Add all rectangles for complete area. Hence, right hand side is twice of area and as initial speed is $0ms^{-1},$ we get, $v_f^2=2\times\text{Area}$
A: The work done (per unit mass) after it has traveled a distance $x$ is the area under the acceleration curve between $0$ and $x$. $$W = 6 x - \frac{x^2}{10}$$
This work goes into kinetic energy (per unit mass) which is $K=\frac{1}{2} v^2$. Equating the two will give the velocity as a function of position
$$ v(x) = \sqrt{12 x - \frac{x^2}{5}} $$
Appendix
Area under acceleration curve $a(x)  = 6 \left(1-\frac{x}{30}\right)$

Area of rectangle$x \, a(x) = 6x-\frac{x^2}{5}$
Area of triangle $\frac{1}{2} x (6-a(x)) = \frac{x^2}{10}$
Total Area $ W = 6x-\frac{x^2}{5} + \frac{x^2}{10} = 6x-\frac{x^2}{10} $
A: According to the graph, accelration is linearly dependent on displacement. Now, assuming the motion is rectillinear(along a straight line, and quite justified for 30 m displacement of a train).
It is not very difficult to find the equation of this straight line (intercept form).
And then, you'll need some calculus to find velocity $\nu$ in terms of displacement $x$, which involves the following substitution : $$a=\nu\frac{d\nu}{dx}$$
Now transfer $dx$ to the other side and integrate. Try and see if it helps.
