Give example of a motion where $x > 0$, $v 0$ at a particular instant.

Give example of a motion where x > 0, v < 0, a > 0 – Class 11 Physics

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x > 0

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Concept understanding — Instantaneous Velocity

Instantaneous Velocity: From "How Fast" to "How Fast Right Now"

You already know average velocity. If a car travels 120 km in 2 hours, its average velocity is 60 km/h. That tells you the overall rate, but it hides everything that happened in between — the traffic jams, the sudden bursts of speed, the moments the car was completely stopped.

Now imagine you want to know the car's velocity at exactly 10:15 AM, not averaged over an hour or a minute. That's instantaneous velocity — the velocity at a single instant of time.

The Intuition: Zooming In

Think of a speedometer needle. When you drive, the needle doesn't stay fixed at 60 km/h. It jumps up when you accelerate, drops when you brake. At any given moment, the needle points to a specific number. That number is your instantaneous speed (velocity, if direction matters).

But here's the puzzle: at a single instant, the car hasn't moved any distance. How can you have a speed if $\Delta t = 0$ ? You can't divide by zero.

The trick is to shrink the time interval smaller and smaller, and see what the average velocity approaches.

The Precise Definition

Let $s(t)$ be the position of an object at time $t$ . The average velocity over a time interval $[t, t+h]$ is:

v_{\text{avg}} = \frac{s(t+h) - s(t)}{h}

Now, let $h$ get closer and closer to 0 (but never equal to 0). If the average velocity settles down to a single number as $h \to 0$ , that number is the instantaneous velocity at time $t$ :

v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}

v(t) = \lim_{h \to 0} \frac{s(t+h) - s(t)}{h}

This limit is exactly the derivative of position with respect to time. In calculus notation: $v(t) = s'(t)$ .

A Concrete Example

Suppose a ball is dropped from rest, and its height (in meters) after $t$ seconds is $s(t) = 4.9t^2$ (ignoring air resistance).

Average velocity from $t=2$ to $t=2.1$ seconds:

v_{\text{avg}} = \frac{4.9(2.1)^2 - 4.9(2)^2}{0.1} = \frac{4.9(4.41 - 4)}{0.1} = \frac{4.9 \times 0.41}{0.1} = 20.09 \text{ m/s}

Average velocity from $t=2$ to $t=2.01$ :

v_{\text{avg}} = \frac{4.9(2.01)^2 - 4.9(2)^2}{0.01} = \frac{4.9(4.0401 - 4)}{0.01} = 19.649 \text{ m/s}

Average velocity from $t=2$ to $t=2.001$ :

v_{\text{avg}} = \frac{4.9(2.001)^2 - 4.9(2)^2}{0.001} = 19.6049 \text{ m/s}

The numbers are converging to 19.6 m/s. That's the instantaneous velocity at $t=2$ seconds.

Using the derivative: $v(t) = 9.8t$ , so $v(2) = 19.6$ m/s. Matches perfectly.

Key Takeaways for Exams

Concept	Meaning	Formula
Average velocity	Total displacement ÷ total time	$\frac{\Delta s}{\Delta t}$
Instantaneous velocity	Velocity at a single moment	$\lim_{h \to 0} \frac{s(t+h)-s(t)}{h}$

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Throw a ball vertically upward and, on the way up, take downward as positive with the origin at the highest point it will reach. While still rising the ball has $x > 0$ (below that origin), $v < 0$ (moving up), and $a = +g > 0$ (gravity, downward).

What the three signs demand

$x$ , $v$ and $a$ are independent at any instant. To have $x>0$ , $v<0$ and $a>0$ together, the object must be at a positive position, moving in the negative direction, while its acceleration points in the positive direction - i.e. the acceleration opposes the velocity, so the object is slowing down (decelerating) as it moves the negative way.

A concrete example

Throw a ball straight up. Choose the coordinate frame:

origin ( $x=0$ ) at the highest point the ball reaches,
downward taken as the positive direction.

Consider an instant while the ball is still on its way up:

it lies below the highest point, so $x > 0$ (positive, since down is positive);
it is moving upward, i.e. in the negative direction, so $v < 0$ ;
gravity acts downward, the positive direction, so $a = +g > 0$ . …

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Concept: Constructing an Explicit Kinematic Function to Satisfy Three Independent Sign Conditions

Method: Build a Concrete Quadratic $x(t)$ and Verify by Direct Substitution — a Braking Reversing Car, Not a Vertical Throw

The stored answer uses a ball thrown vertically upward (with a chosen origin and sign convention) to satisfy $x>0,\,v<0,\,a>0$ . This method instead constructs a purely horizontal scenario — a car backing out of a driveway while braking — with an explicit numeric quadratic position function, and confirms all three conditions by plugging in a number, rather than by choosing a coordinate convention around gravity.

What the three signs require, restated

$x>0$ : the object is on the positive side of the origin. $v<0$ : it is moving in the negative direction. $a>0$ : its acceleration points in the positive direction — i.e., opposing the (negative) velocity, so the object is decelerating while moving backward.

A concrete scenario and function

Scenario: a car reversing out of a driveway. Let $x=0$ be a fixed reference point on the driveway, positive $x$ pointing into the driveway (away from the road). The car starts well inside the driveway, at $x=10\ \text{m}$ , and reverses out (moving in the $-x$ direction) while the driver brakes — so the car's speed is decreasing as it moves backward, meaning its acceleration points in $+x$ (opposing the backward motion).
Write an explicit quadratic position function consistent with this description (uniform deceleration, i.e. constant positive $a$ ):

x(t) = 10 - 8t + 2t^2 \quad (\text{SI units, } t\ge0)

Differentiate to get velocity and acceleration:

v(t) = \frac{dx}{dt} = -8+4t, \qquad a(t) = \frac{dv}{dt} = 4

Acceleration is a constant $+4\ \text{m s}^{-2}$ — always positive, matching the "braking while reversing" description (constant, since gentle steady braking is being modelled).

Evaluate all three functions at $t=1\ \text{s}$ :

x(1) = 10-8+2 = 4\ \text{m} \;(>0), \qquad v(1) = -8+4 = -4\ \text{m s}^{-1}\;(<0), \qquad a(1) = 4\ \text{m s}^{-2}\;(>0)

All three conditions hold simultaneously at $t=1\ \text{s}$ : $x>0$ , $v<0$ , $a>0$ — confirmed by direct substitution into an explicit, self-consistent kinematic function, with no coordinate-convention choice needed (the car's own physical position, without redefining "up" or "down"). …

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Q.Give example of a motion where $x > 0$ , $v < 0$ , $a > 0$ at a particular instant.

Concept understanding — Instantaneous Velocity

Instantaneous Velocity: From "How Fast" to "How Fast Right Now"

The Intuition: Zooming In

The Precise Definition

A Concrete Example

Key Takeaways for Exams

Motion with $x > 0$ , $v < 0$ , $a > 0$

What the three signs demand

A concrete example

Concept: Constructing an Explicit Kinematic Function to Satisfy Three Independent Sign Conditions

What the three signs require, restated

A concrete scenario and function

Q.Give example of a motion where x>0x > 0x>0, v<0v < 0v<0, a>0a > 0a>0 at a particular instant.

What the three signs demand

A concrete example

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Q.Give example of a motion where $x > 0$ , $v < 0$ , $a > 0$ at a particular instant.

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