2.2.3 Zero Initial Velocity

If a (rectilinear) uniform acceleration motion starts from rest, then the SUVAT equations can be further simplified as follow:

$\displaystyle \begin{aligned}v&=u+at\text{ }\Rightarrow \text{ }v=at\\s&=ut+\frac{1}{2}a{{t}^{2}}\text{ }\Rightarrow \text{ }s=\frac{1}{2}a{{t}^{2}}\\{{v}^{2}}&={{u}^{2}}+2as\text{ }\Rightarrow \text{ }v=\sqrt{{2a}}\sqrt{s}\end{aligned}$

v-t graph

The v-t graph is a straight line graph $\displaystyle (v=at)$ passing through the origin (because $\displaystyle u=0$ ). The gradient of the graph corresponds to the constant acceleration a.

The area under the graph corresponds to the displacement s.

By counting the triangles that make up the area under the graph, we can deduce that the distance travelled during the 1^st, 2^nd, 3^rd, 4^th … unit of time is 1, 3, 5, 7 … units of distance. This arithmetic progression makes sense since the velocity increases linearly with time.

s-t graph

The s-t graph is a quadratic curve $\displaystyle (s=\frac{1}{2}a{{t}^{2}})$ . As such, the distance travelled after 1, 2, 3, … units of time are 1, 4, 9, 16, … units of distance.

Note that the gradient of the graph at any point corresponds to v, the velocity at that time instant.

As such, the graph must be drawn with zero gradient at $\displaystyle t=0$ since $\displaystyle u=0$ .

v-s graph

The v-s graph is a square-root function $\displaystyle (v=\sqrt{{2as}})$ . So, v does increase with s, but at decreasing rate. Do note that the v-s graph has a vertical gradient at $\displaystyle s=0$ , because that’s what a square-root graph looks like.