A particle located at x = 0 at time t = 0, starts moving along the positive x-direction with a veloc

Q: Why did we replace $v$ with $dx/dt$?

Velocity is defined as the rate of change of position. Replacing $v$ with $dx/dt$ introduces the variables $x$ and $t$ into the equation, allowing us to find the relationship between them.

Q: How did $\sqrt{x}$ become $x^{-1/2}$?

By the laws of exponents, $\sqrt{x}$ is $x^{1/2}$. When we move it from the denominator to the numerator, the exponent becomes negative, resulting in $x^{-1/2}$.

Q: Why do we use definite integrals instead of adding a constant $C$?

Definite integrals automatically incorporate the initial conditions ($x=0$ at $t=0$) as the lower limits, which is often faster and less prone to algebraic errors than finding the constant of integration $C$.

Q: What does the result $x \propto t^2$ imply about the particle's motion?

From kinematics, we know that if displacement is proportional to the square of time ($x = \frac{1}{2}at^2$), the particle is moving with a constant acceleration.

Q: Could we have solved this by finding acceleration first?

Yes! You could find acceleration using $a = v \frac{dv}{dx}$. Since $v = \alpha \sqrt{x}$, $dv/dx = \frac{\alpha}{2\sqrt{x}}$. Multiplying them gives $a = \frac{\alpha^2}{2}$, which is constant. Then use $x = \frac{1}{2}at^2$.

Question

A particle located at $x = 0$ at time $t = 0$, starts moving along the positive x-direction with a velocity $v$ that varies as $v = \alpha \sqrt{x}$. The displacement of the particle varies with time as

Options:

(A) $t^2$

(B) $t$

(C) $t^{1/2}$

(D) $t^3$

Sigma Percentile · Accepted Answer

### The Dynamic World of Variable Velocity

Imagine you are observing a particle that just started its journey. It begins exactly at the origin, where $x = 0$, and the stopwatch has just been clicked, so $t = 0$.

But this isn't your standard uniform motion. The particle is accelerating, and its velocity is governed by a peculiar rule: $v = \alpha \sqrt{x}$.

This means the further the particle travels, the faster it goes! Our mission is to uncover the hidden relationship between the particle's displacement $x$ and the time $t$.

### Bridging Kinematics and Calculus

To solve this, we need a mathematical bridge. We know that velocity is not just a standalone quantity; it is the fundamental rate at which position changes over time.

**Instantaneous velocity** is mathematically defined as the derivative of position with respect to time.

$$v = \frac{dx}{dt}$$

By substituting this definition into our given velocity function, we transform a simple algebraic equation into a powerful **differential equation**.

$$\frac{dx}{dt} = \alpha \sqrt{x}$$

This equation is the key to unlocking the particle's entire future.

### The Art of Separation and Integration

To solve this differential equation, we must use a technique called **separation of variables**. We want all the $x$ terms on one side and all the $t$ terms on the other.

By dividing both sides by $\sqrt{x}$ and multiplying by $dt$, we achieve this separation.

$$\frac{dx}{\sqrt{x}} = \alpha dt$$

Now, the stage is set for integration. We will integrate both sides, using our initial conditions as the lower bounds. The particle starts at $x = 0$ when $t = 0$, and we want to find its position $x$ at any general time $t$.

$$\int_0^x x^{-1/2} dx = \int_0^t \alpha dt$$

### Executing the Math

Let's perform the integration. The power rule for integration tells us to add one to the exponent and divide by the new exponent.

$$\left[ \frac{x^{1/2}}{1/2} \right]_0^x = \alpha [t]_0^t$$

Substituting the limits, we get a clean algebraic relation.

$$2\sqrt{x} = \alpha t$$

### The Final Revelation

We are almost there! To isolate the displacement $x$, we simply square both sides of the equation.

$$4x = \alpha^2 t^2$$

$$x = \frac{\alpha^2}{4} t^2$$

Since $\alpha$ is a given constant, the term $\frac{\alpha^2}{4}$ is also a constant. This leads us to our grand conclusion.

$$x \propto t^2$$

The displacement of the particle is directly proportional to the square of the time elapsed. This is the hallmark of motion with constant acceleration, a beautiful result hidden within a variable velocity equation!

Animated Solution for Physics - Kinematics: A particle located at $x = 0$ at time $t = 0$ , starts moving along the positive x-direction with a velocity $v$ that varies as $v = α x$ . The displacement of the particle varies with time as

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Visualized Solution

$v = α x$

$v = \frac{d x}{d t}$

$\frac{d x}{d t} = α x$

$\frac{d x}{x} = α d t$

$\int_{0}^{x} x^{- 1/2} d x = \int_{0}^{t} α d t$

$2 x = α t$

$x \propto t^{2}$

$a = constant$

The Sigma Insight: Motion in a Straight Line

The Dynamic World of Variable Velocity

Bridging Kinematics and Calculus

The Art of Separation and Integration

Executing the Math

The Final Revelation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=αx​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=dtdx​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtdx​=αx​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x​dx​=αdt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫0x​x−1/2dx=∫0t​αdt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2x​=αt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∝t2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=constant