The position, velocity and acceleration of a particle moving with a constant acceleration can be rep

Q: Why is the position graph a parabola?

Because integrating a linear velocity function ($at+u$) yields a quadratic position function ($\frac{1}{2}at^2+ut+x_0$), which graphically is a parabola.

Q: Can the velocity graph be a curve for constant acceleration?

No, constant acceleration means velocity changes at a constant rate, resulting in a straight line.

Q: What if the initial velocity is zero?

The velocity graph would start from the origin, but it would still be a straight line.

Q: How do we know the parabola opens upwards?

Since acceleration is positive (as seen in the constant positive graph), the coefficient of $t^2$ is positive, making it an upward-opening parabola.

Q: Why is option (d) incorrect?

In option (d), the acceleration is shown as zero, but the velocity is increasing. This is physically impossible since an increasing velocity requires a non-zero acceleration.

Question

The position, velocity and acceleration of a particle moving with a constant acceleration can be represented by

Sigma Percentile · Accepted Answer

### The Essence of Motion Graphs

When a particle moves, its position, velocity, and acceleration are intimately connected through the language of calculus. Graphs are the visual storytelling of these mathematical relationships. In this problem, we are given a simple yet powerful constraint: the particle moves with **constant acceleration**. Let's decode what this means for each of its motion graphs.

### Acceleration: The Foundation

The problem states that the acceleration $a$ is constant. This means that as time $t$ ticks forward, the value of acceleration does not change.

Mathematically, we write this as:
$$a(t) = 	ext{constant}$$

Graphically, a constant value over time is represented by a horizontal straight line. If we assume the acceleration is positive, this line will be parallel to the time axis and above it. This is our first crucial clue.

### Velocity: The Linear Climb

Acceleration is defined as the rate of change of velocity. 
$$a = \frac{dv}{dt}$$

To find the velocity, we integrate the constant acceleration with respect to time:
$$v(t) = \int a \, dt = at + u$$
where $u$ is the initial velocity.

This equation, $v(t) = at + u$, is a linear equation in the form $y = mx + c$. The slope of this line is the acceleration $a$, which is constant and positive. Therefore, the velocity-time graph must be a **straight line with a positive slope**.

### Position: The Parabolic Arc

Finally, let's determine the position. Velocity is the rate of change of position.
$$v = \frac{dx}{dt}$$

To find the position, we integrate the linear velocity function:
$$x(t) = \int v \, dt = \int (at + u) \, dt$$
$$x(t) = \frac{1}{2}at^2 + ut + x_0$$
where $x_0$ is the initial position.

This is a quadratic equation in terms of time $t$. In geometry, a quadratic equation represents a **parabola**. Since the coefficient of $t^2$ (which is $\frac{1}{2}a$) is positive, this parabola opens upwards.

### Bringing It All Together

We have deduced the shapes of all three graphs:
1.  **Acceleration-time:** A horizontal straight line.
2.  **Velocity-time:** A straight line with a positive slope.
3.  **Position-time:** An upward-opening parabola.

Looking at the given options, only **Option (b)** perfectly matches all three of these deduced shapes. The beauty of kinematics lies in this perfect synchronization between calculus and geometry!

Animated Solution for Physics - Kinematics: The position, velocity and acceleration of a particle moving with a constant acceleration can be represented by

Select Answer:

Visualized Solution

$a = constant$

$a (t)$

$v (t)$

$x (t)$

$Conclusion$

$What if a < 0 ?$

The Sigma Insight: Motion Graphs

The Essence of Motion Graphs

Acceleration

Velocity

Position

Bringing It All Together

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