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:

We are told that the position (displacement) of the car at time $$t$$ is $$f(t)=at^{2}+bt+c$$ with $$a,b,c>1$$ and $$t>0$$. We wish to know at which point inside the interval $$[t_{1},t_{2}]$$ the average speed (average velocity) is actually attained.

First, recall the result from basic calculus called the Mean Value Theorem (MVT) for derivatives: if a function $$f$$ is continuous on $$[t_{1},t_{2}]$$ and differentiable on $$(t_{1},t_{2})$$, then there exists some number $$\xi$$ in the open interval $$(t_{1},t_{2})$$ such that

$$f'(\xi)=\frac{f(t_{2})-f(t_{1})}{\,t_{2}-t_{1}\,}.$$

The right-hand side of this equality is the average rate of change of $$f$$ over $$[t_{1},t_{2}]$$, which, for a displacement function, is the average velocity (average speed in magnitude, because the motion is on a straight line). Hence, by the MVT, the instantaneous velocity at some interior point $$\xi$$ equals this average velocity.

To locate that $$\xi$$ explicitly, we must compute both the derivative $$f'(t)$$ and the average velocity.

We differentiate $$f(t)=at^{2}+bt+c$$ with respect to $$t$$. Using the power rule $$\frac{d}{dt}(t^{n})=nt^{\,n-1}$$, we obtain

$$f'(t)=2at+b.$$

Next we calculate the average velocity (average speed) over $$[t_{1},t_{2}]$$:

$$\frac{f(t_{2})-f(t_{1})}{t_{2}-t_{1}} =\frac{\,\bigl(a\,t_{2}^{2}+b\,t_{2}+c\bigr)-\bigl(a\,t_{1}^{2}+b\,t_{1}+c\bigr)\,}{t_{2}-t_{1}}.$$

The constants $$c$$ cancel, leaving

$$\frac{a\,(t_{2}^{2}-t_{1}^{2})+b\,(t_{2}-t_{1})}{t_{2}-t_{1}}.$$

Now, recall the algebraic identity $$t_{2}^{2}-t_{1}^{2}=(t_{2}-t_{1})(t_{2}+t_{1}).$$ Substituting this into the numerator, we get

$$\frac{a\,(t_{2}-t_{1})(t_{2}+t_{1}) + b\,(t_{2}-t_{1})}{t_{2}-t_{1}}.$$

Since a common factor $$t_{2}-t_{1}$$ appears in every term in the numerator, it cancels with the denominator, yielding

$$a\,(t_{1}+t_{2}) + b.$$

Thus the average velocity is $$a(t_{1}+t_{2}) + b.$$

By the Mean Value Theorem, we must have some $$\xi \in (t_{1},t_{2})$$ satisfying

$$f'(\xi)=2a\,\xi + b = a(t_{1}+t_{2}) + b.$$

Subtracting $$b$$ from both sides gives

$$2a\,\xi = a\,(t_{1}+t_{2}).$$

Because $$a>0$$, we can divide both sides by $$2a$$:

$$\xi=\frac{t_{1}+t_{2}}{2}.$$

This point $$\xi$$ equals the arithmetic mean of $$t_{1}$$ and $$t_{2}$$. Looking at the listed choices, this corresponds exactly to Option C:

$$\frac{(t_{1}+t_{2})}{2}.$$

Hence, the correct answer is Option C.