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For motion in a straight line with constant acceleration, we derived two primary equations given by Eq. (4.4a) and (4.4b). Using these two equations, three more can be derived, of which we derived one (4.4c). Derive the remaining two equations: \(s = vt - \tfrac{1}{2}at^2\) and \(s = \tfrac{1}{2}(u+v)t\). Using the area of a trapezium, derive the second equation above.
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Trapezium area: the velocity-time graph is a trapezium with parallel sides \(u\) and \(v\) and width \(t\): $$s = \tfrac{1}{2}(u+v)\,t$$
Other equation: from \(v = u + at\) we get \(u = v - at\). Substituting into \(s = ut + \tfrac{1}{2}at^2\): \( s = (v-at)t + \tfrac{1}{2}at^2 = vt - at^2 + \tfrac{1}{2}at^2 \), so $$s = vt - \tfrac{1}{2}at^2$$
