A particle is moving with speed v = b\sqrt{x} along positive X-axis. Calculate the speed of the part

Q: Why did we differentiate with respect to time $t$ instead of position $x$?

We differentiated with respect to time because the question asks for the speed at a specific time $t = \tau$. Finding the acceleration $a = \frac{dv}{dt}$ helps us connect the velocity to time using kinematic equations.

Q: Can I use the formula $v^2 = u^2 + 2ax$ here?

Yes, you can! However, that formula will give you the velocity at a specific position $x$. Since the question asks for velocity at a specific time $t$, using $v = u + at$ is much more direct.

Q: What if the acceleration we found wasn't constant?

If the acceleration turned out to be a function of time or position (not a constant), we could not use the standard equations of motion like $v = u + at$. We would have to rely entirely on integration to find the velocity as a function of time.

Q: How does the alternative integration method work?

By writing $v$ as $\frac{dx}{dt}$, you get $\frac{dx}{dt} = b\sqrt{x}$. You can separate the variables to get $\frac{dx}{\sqrt{x}} = b \, dt$. Integrating both sides gives $2\sqrt{x} = bt$. You can then substitute $\sqrt{x} = \frac{bt}{2}$ back into the original equation $v = b\sqrt{x}$ to get $v = \frac{b^2t}{2}$.

Question

A particle is moving with speed $v = b\sqrt{x}$ along positive X-axis. Calculate the speed of the particle at time $t = 	au$ (assume that the particle is at origin at $t = 0$).

Options:

(A) $\frac{b^2	au}{4}$

(B) $\frac{b^2	au}{2}$

(C) $b^2	au$

(D) $\frac{b^2	au}{\sqrt{2}}$

Sigma Percentile · Accepted Answer

### The Position-Dependent Velocity

Imagine a particle moving along the positive x-axis. Its speed isn't a simple constant; it dynamically changes based on its position $x$. The relationship governing this motion is given by the equation:

$$v = b\sqrt{x}$$

Our goal is to find the speed of this particle at a specific time $t = 	au$. The challenge here is that our velocity is given in terms of position ($x$), but we need it in terms of time ($t$). How do we bridge this gap?

### The Master Strategy: Finding Acceleration

To transition from a position-dependent velocity to a time-dependent one, finding the acceleration is a brilliant strategy. Acceleration is the rate of change of velocity with respect to time. Let's differentiate our velocity equation with respect to time $t$:

$$a = \frac{dv}{dt} = \frac{d}{dt}(b\sqrt{x})$$

Here is where we must be careful. Since we are differentiating a function of $x$ with respect to $t$, we must apply the **chain rule**. The derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$, but we must multiply it by the derivative of the inner function, which is $\frac{dx}{dt}$.

$$a = b \cdot \frac{1}{2\sqrt{x}} \cdot \frac{dx}{dt}$$

### The Calculus Magic

Now, look closely at the term $\frac{dx}{dt}$. What does it represent physically? It is simply the velocity $v$ of the particle! Let's substitute $v$ back into our acceleration equation:

$$a = \frac{b}{2\sqrt{x}} \cdot v$$

We already know that $v = b\sqrt{x}$. When we plug this expression into our equation, something beautiful happens:

$$a = \frac{b}{2\sqrt{x}} \cdot (b\sqrt{x})$$

The $\sqrt{x}$ terms perfectly cancel each other out! We are left with:

$$a = \frac{b^2}{2}$$

This is a profound result. It tells us that despite the velocity changing with position, the **acceleration of the particle is strictly constant**.

### The Kinematics Finale

Since the acceleration is constant, we are now in the familiar territory of standard kinematics. We can use the first equation of motion:

$$v = u + at$$

Before we plug in the values, we need the initial velocity $u$. The problem states that at $t = 0$, the particle is at the origin ($x = 0$). Plugging $x = 0$ into our original velocity equation gives:

$$u = b\sqrt{0} = 0$$

Now, we have everything we need. Let's substitute $u = 0$, $a = \frac{b^2}{2}$, and $t = 	au$ into our equation of motion:

$$v = 0 + \left(\frac{b^2}{2}ight)	au$$

$$v = \frac{b^2	au}{2}$$

And there we have it! The speed of the particle at time $	au$ is exactly $\frac{b^2	au}{2}$.

### The Alternate Path: Direct Integration

Could we have solved this without finding the acceleration? Absolutely! We can write $v$ as $\frac{dx}{dt}$ and solve the resulting differential equation by separating the variables:

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

$$\int x^{-1/2} dx = \int b \, dt$$

Integrating both sides yields $2\sqrt{x} = bt$, which means $\sqrt{x} = \frac{bt}{2}$. Substituting this back into $v = b\sqrt{x}$ gives $v = b\left(\frac{bt}{2}ight) = \frac{b^2t}{2}$. Both paths lead to the same elegant truth!

Animated Solution for Physics - Kinematics: A particle is moving with speed $v = b x$ along positive X-axis. Calculate the speed of the particle at time $t = τ$ (assume that the particle is at origin at $t = 0$ ).

Select Answer:

Visualized Solution

$v (x)$ Relationship

Finding Acceleration $a$

Chain Rule Application

Substituting $v$

Constant Acceleration

Initial Velocity $u$

Equation of Motion

Final Calculation

Alternative Method

The Sigma Insight: Motion in a Straight Line

The Position-Dependent Velocity

The Master Strategy

The Calculus Magic

The Kinematics Finale

The Alternate Path

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