The position of a moving car at time t is giv | vedclass

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Question

(D) The average speed of the car over the interval $[t_{1}, t_{2}]$ is given by the formula: $	ext{Average Speed} = \frac{f(t_{2}) - f(t_{1})}{t_{2}

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$\frac{t_{1} + t_{2}}{2}$

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(D) The average speed of the car over the interval $[t_{1}, t_{2}]$ is given by the formula: $	ext{Average Speed} = \frac{f(t_{2}) - f(t_{1})}{t_{2} - t_{1}}$Substituting $f(t) = at^{2} + bt + c$ into the formula:$\frac{(at_{2}^{2} + bt_{2} + c) - (at_{1}^{2} + bt_{1} + c)}{t_{2} - t_{1}} = \frac{a(t_{2}^{2} - t_{1}^{2}) + b(t_{2} - t_{1})}{t_{2} - t_{1}}$$= \frac{a(t_{2} - t_{1})(t_{2} + t_{1}) + b(t_{2} - t_{1})}{t_{2} - t_{1}} = a(t_{1} + t_{2}) + b$We want to find the time $t$ such that the instantaneous speed $f'(t)$ equals this average speed.$f'(t) = \frac{d}{dt}(at^{2} + bt + c) = 2at + b$Equating the instantaneous speed to the average speed:$2at + b = a(t_{1} + t_{2}) + b$$2at = a(t_{1} + t_{2})$$t = \frac{t_{1} + t_{2}}{2}$

The position of a moving car at time $t$ is given by $f(t) = at^{2} + bt + c, t > 0,$ where $a, b,$ and $c$ are real numbers greater than $1.$ Then the average speed of the car over the time interval $[t_{1}, t_{2}]$ is attained at the point

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