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Prove a property of a particle moving according to a given function

A particle moves along a straight line and has position at time t given by the function f(t). The initial velocity of the particle is f'(0) = 0 and the acceleration function is continuous, given by f''(t) \geq 6 for all t \in [0,1]. Prove that f'(t) \geq 3 for all t in some interval [a,b] where

    \[ 0 \leq a < b \leq 1, \qquad \text{and} \qquad b-a = \frac{1}{2}. \]


Proof. From the second fundamental theorem of calculus, we have

    \begin{align*}  &&f'(t) - f'(0) &= \int_0^t f''(x) \, dx \\  \implies && f'(t) &\geq \int_0^t 6 \, dx \\  \implies && f'(t) &\geq 6t \\  \implies && f' \left( \frac{1}{2} \right) &\geq 3. \end{align*}

Then, since f''(x) \geq 6 implies f''(x) > 0 implies f'(x) is increasing for all t \in [0,1], we have

    \[ f'(t) \geq 3 \qquad \text{for all } t \in \left[ \frac{1}{2}, 1 \right]. \]

Now, let b = 1 and a = \frac{1}{2} and we have f'(t) \geq 3 for all t \in [a,b] where 0 \leq a < b \leq 1 and b - a = \frac{1}{2}, as requested. \qquad \blacksquare

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