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 $. \qquad \blacksquare$

One comment

January 11, 2023 at 7:29 pm

William says:

Would this proof still work for a strict inequality (continuous acceleration f”(t) >= 6 for all t in the interval 0 < t < 1) or do we need to include continuity on t = 0 and t = 1?

Stumbling Robot

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

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