Deriving displacement via double integration of v-t graph gradient

Steps:

• The gradient of the v-t graph is constant acceleration $a = \frac{dv}{dt}$.
• Integrate once with respect to time: $v = u + at$.
• Integrate velocity with respect to time: $s = \int_0^t v \, dt = \int_0^t (u + at) \, dt = ut + \frac{1}{2}at^2$.

Why C is correct:

• C follows from successive time integrations of constant acceleration, defined as the v-t graph gradient, per the fundamental theorem of calculus.

Why the others are wrong:

• A: Relies on geometric area under the v-t graph (trapezoid formula), beyond just the gradient.
• B: Comes directly from the single integration of the gradient but omits displacement.
• D: Uses chain rule integration $v \, dv = a \, ds$, incorporating displacement elements not solely from the time gradient.

Final answer: C