In a speed vs. time graph, the area under the curve gives you the distance, since the distance is integral vdt. I hope you know this. The accn is dv/dt, and so the slope of the tangent at a point represents the accn.
Look at the start: the v is -5 m/s for some time t1 (not specified). So, the particle is moving left (say), and the dist covered will be the product of -5*t1 (in metres). It’s negative.
Next, at a single instant, the speed becomes +5 from -5, showing that accn is infinitely large and not defined at that instant. Typically, this represents an impulsive force acting for a very short time (what we call collisions).
Then the speed decreases linearly from 5 to 0, and the accn is uniform because the slope of the graph is constant. The accn a= (0-5)/t2 m/s/s. The dist is the area of the triangle, giving the formula d= ½ v0*t2.
Then the speed is zero => dist covered is 0 and accn =0.
You get the drift, I hope. At the end, you can sum up all the distances, with proper signs, to find the total dist travelled. Find the accns for each of the segments by finding the slope.
