The position of a moving car at time t is given by f(t) = at² + bt + c, t 0, where a, b and c are

Q: Wait, why do we reject the negative root here?

Actually, in this specific quadratic case, we don't have a quadratic equation in $t$ to solve! When we equate the instantaneous speed to the average speed, the $t^2$ terms cancel out perfectly, leaving us with a linear equation. There is only one unique solution, so there are no roots to reject.

Q: Why can't we just use energy conservation?

Energy conservation relates to forces and work, but here we are dealing with pure kinematics—the geometry of motion. We are looking for a specific point in time where rates of change align, which is a problem of calculus and geometry, not dynamics.

Q: Is this related to the Mean Value Theorem?

Absolutely! You have just derived a specific case of Lagrange's Mean Value Theorem (LMVT). LMVT states that for a continuous and differentiable function, there exists at least one point $c$ in $(t_1, t_2)$ where the instantaneous rate equals the average rate. For parabolas, that point is always the midpoint.

Question

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:

Options:

(A) $2a (t_1+t_2) + b$

(B) $(t_2-t_1)/2$

(C) $a (t_2-t_1) + b$

(D) $(t_1+t_2)/2$

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Motion

Welcome, future engineer. Today, we aren't just solving a kinematics problem; we are uncovering a hidden symmetry of nature.

When you look at the equation $f(t) = at^2 + bt + c$, don't just see a collection of variables. See a story. This is the story of a car accelerating, its position changing in a beautiful, parabolic arc.

We are going to explore the relationship between the 'average' of a journey and the 'instant' of a moment.

### Phase 1: The Geometry of the Journey

Imagine you are standing on the side of the road, watching this car. You mark its position at two times: $t_1$ and $t_2$.

The average speed over this interval is not just a number; it is the slope of the secant line connecting the point $(t_1, f(t_1))$ to $(t_2, f(t_2))$. Mathematically, we define this as:

$$	ext{Average Speed} = \frac{f(t_2) - f(t_1)}{t_2 - t_1}$$

This is the 'big picture' view. It tells us how fast the car was moving on average, ignoring the fluctuations in between.

### Phase 2: The Precision of the Instant

Now, shift your focus. What if you want to know the speed at one specific, frozen moment in time? That is the instantaneous speed.

In the language of calculus, this is the derivative of the position function. Let's take our function $f(t) = at^2 + bt + c$ and find its rate of change:

$$f'(t) = \frac{d}{dt}(at^2 + bt + c) = 2at + b$$

This expression, $2at + b$, is the 'pulse' of the car. It tells us exactly how fast the car is moving at any time $t$.

### Phase 3: The Algebraic Dance

Now, let's perform the calculation. We need to find the displacement $f(t_2) - f(t_1)$. Watch how the constants behave:

$$f(t_2) - f(t_1) = (at_2^2 + bt_2 + c) - (at_1^2 + bt_1 + c)$$

Notice that the constant $c$ vanishes. It doesn't matter where the car started relative to the origin; only the change matters. We are left with:

$$a(t_2^2 - t_1^2) + b(t_2 - t_1)$$

Using the difference of squares identity, $t_2^2 - t_1^2 = (t_2 - t_1)(t_2 + t_1)$, we can factor out the $(t_2 - t_1)$ term:

$$f(t_2) - f(t_1) = (t_2 - t_1)[a(t_1 + t_2) + b]$$

When we divide this by the total time $(t_2 - t_1)$ to get the average speed, the term cancels out beautifully, leaving us with:

$$	ext{Average Speed} = a(t_1 + t_2) + b$$

### Phase 4: The Moment of Truth

The problem asks us to find the time $t$ where the instantaneous speed equals this average speed. We set our two expressions equal:

$$2at + b = a(t_1 + t_2) + b$$

Subtract $b$ from both sides, and we get $2at = a(t_1 + t_2)$. Since $a 
eq 0$, we can safely divide by $a$:

$$2t = t_1 + t_2 \Rightarrow t = \frac{t_1 + t_2}{2}$$

### The Takeaway

Look at that result: $t = \frac{t_1 + t_2}{2}$. It is the exact midpoint of the interval.

This isn't a coincidence. You have just proven a fundamental property of parabolas: the tangent line is parallel to the secant line exactly at the midpoint of the interval.

This is the essence of **Lagrange's Mean Value Theorem** in action. You have moved beyond just solving for $t$; you have understood the geometric soul of the quadratic function. Keep this intuition with you—it will serve you well in your journey through physics.

Animated Solution for Mathematics - Differentiation: The position of a moving car at time $t$ is given by $f (t) = a t^{2} + b t + c, t > 0$ , 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:

Select Answer:

Visualized Solution

Position Function $f (t)$

Instantaneous Speed $v (t)$

Average Speed Concept

Calculating Displacement

Simplifying the Expression

Final Average Speed

Equating the Speeds

Solving for $t$

Key Takeaway

The Sigma Insight: Mean Value Theorems

The Hidden Symmetry of Motion

Phase 1

Phase 2

Phase 3

Phase 4

The Takeaway

Similar Questions

Let $f$ be any function continuous on $[a, b]$ and twice differentiable on $(a, b)$ . If for all $x \in (a, b)$ , $f^{'} (x) > 0$ and $f^{''} (x) < 0$ , then for any $c \in (a, b)$ , $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than:

If $f (x)$ is twice differentiable and continuous function in $x \in [a, b]$ also $f^{'} (x) > 0$ and $f^{''} (x) < 0$ and $c \in (a, b)$ then $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than

If the function $f : [0, 4] \to R$ is differentiable then show that (i) For $a, b \in (0, 4), (f (4))^{2} - (f (0))^{2} = 8 f^{'} (a) f (b)$

Let the function, $f : [- 7, 0] \to R$ be continuous on $[- 7, 0]$ and differentiable on $(- 7, 0)$ . If $f (- 7) = - 3$ and $f^{'} (x) \leq 2$ , for all $x \in (- 7, 0)$ , then for all such functions $f$ , $f (- 1) + f (0)$ lies in the interval:

Let $S$ be the set of all functions $f : [0, 1] \to R$ , which are continuous on $[0, 1]$ and differentiable on $(0, 1)$ . Then for every $f$ in $S$ , there exists a $c \in (0, 1)$ , depending on $f$ , such that:

Let $g : R \to R$ be a non constant twice differentiable such that $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ . If a real valued function $f$ is defined as $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ , then

If $2 a + 3 b + 6 c = 0$ , then at least one root of the equation $a x^{2} + b x + c = 0$ lies in the interval

If $a + b + c = 0$ , then the quadratic equation $3 a x^{2} + 2 b x + c = 0$ has

Let the function, $f : [- 7, 0] \to R$ be continuous on $[- 7, 0]$ and differentiable on $(- 7, 0)$ . If $f (- 7) = - 3$ and $f^{'} (x) \leq 2$ , for $x \in (- 7, 0)$ , then for all such functions $f, f (- 1) + f (0)$ lies in the interval:

For all twice differentiable functions $f : R \to R$ , with $f (0) = f (1) = f^{'} (0) = 0$