The velocity-time graph of a particle moving along a straight line is a straight line inclined at an angle of 45^circ to the time axis. The displacem…

The slope of the v-t graph gives acceleration, so $a = \tan(45^\circ) = 1\text{ m/s}^2$. Initial velocity can be assumed to be 0 if not specified. Displacement $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(1)(4)^2 = 8\text{ m}$. Alternatively, area under the v-t graph = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8\text{ m}$.