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

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

Concept: Instantaneous Velocity and Acceleration — position, velocity and acceleration are independent at a given instant; we need $x>0$, $v<0$ (moving toward the origin) and $a>0$ (acceleration opposing the velocity, i.e. the object is decelerating).

Example: A ball thrown vertically upward, examined while it is still rising. Choose the origin at the ball's highest point and take downward as positive.

• While rising, the ball is below the highest point, so its position (measured downward from the top) is $x > 0$.
• It is moving upward, i.e. in the negative direction, so $v < 0$. …

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 $a = +g > 0$. …

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

1. 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).

2. Write an explicit quadratic position function consistent with this description (uniform deceleration, i.e. constant positive $a$):

3. Differentiate to get velocity and acceleration:

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).

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

5. 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"). …