For the same journey (uniform deceleration from 40 to 25 ms$^{-1}$ over the first 30 s, then constant 25 ms$^{-1}$ for the next 500 m), which statement correctly describes the shape of the displacement-time $(t,x)$ graph?

The gradient of a $(t,x)$ graph is the velocity. During $0\le t\le30$ the velocity decreases from 40 to 25 ms$^{-1}$, so the gradient of the graph decreases, which is a concave-down curve. For $t>30$ the velocity is constant at 25 ms$^{-1}$, so the graph becomes a straight line with gradient 25. Having the curved and straight portions the wrong way round misassigns which phase has changing velocity. Getting the sections right but the curvature backwards is wrong because a decreasing velocity gives a concave-down, not concave-up, curve. The velocity is not constant throughout the whole journey, so the displacement cannot be linear in time from start to finish.