If the velocity of a body related to displacement x is given by v = √(5000 + 24x) m/s, then the acce

Q: Why can't I just differentiate $v$ with respect to $t$ directly?

Because the given expression for $v$ is in terms of $x$, not $t$. To differentiate with respect to $t$, you would need to know how $x$ changes with $t$, which requires the chain rule.

Q: What happens if I don't square the equation and differentiate the square root directly?

You will get the exact same answer! $\frac{dv}{dx} = \frac{1}{2\sqrt{5000+24x}} \cdot 24 = \frac{12}{v}$. Then multiplying by $v$ gives $a = v \cdot \frac{12}{v} = 12$. Squaring just saves you from dealing with fractions.

Q: Is the acceleration constant in this problem?

Yes. Because the derivative $v \frac{dv}{dx}$ evaluates to a pure number ($12$), the acceleration does not depend on time or position. It remains constant throughout the motion.

Q: Can I always use the shortcut $v^2 = u^2 + 2ax$?

You can only use this shortcut if the given equation is linear in $x$ when $v$ is squared. If the equation was $v^2 = 5000 + 24x^2$, the acceleration would not be constant, and the kinematic equations would fail.

Q: What does the $5000$ represent physically?

In the context of the kinematic equation $v^2 = u^2 + 2ax$, the constant term $5000$ represents the square of the initial velocity ($u^2$). So, the initial velocity of the body at $x=0$ is $\sqrt{5000}$ m/s.

Question

If the velocity of a body related to displacement $x$ is given by $v = \sqrt{5000 + 24x}$ m/s, then the acceleration of the body is ...... m/s$^2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a particle moving along a straight line. Instead of being given its velocity at a specific time $t$, we are given its velocity at a specific position $x$. The relationship is defined as:

$$v = \sqrt{5000 + 24x}$$

This is a classic kinematic scenario where time is implicitly hidden, and we must extract the acceleration using spatial derivatives.

### The Master Equation

We know that acceleration is the rate of change of velocity with respect to time, $a = \frac{dv}{dt}$. However, our velocity function depends on $x$, not $t$. This is where the **chain rule** comes to our rescue.

We can rewrite the derivative as:

$$a = \frac{dv}{dx} \cdot \frac{dx}{dt}$$

Since the rate of change of position $\frac{dx}{dt}$ is simply the velocity $v$, our master equation becomes:

$$a = v \frac{dv}{dx}$$

This elegant formula is the key to solving any problem where velocity is a function of displacement.

### The Smart Algebraic Move

We could directly differentiate $v = \sqrt{5000 + 24x}$ with respect to $x$. The derivative of a square root involves fractions and can be slightly messy.

Instead, let's make a smart algebraic move. By squaring both sides of the equation, we completely eliminate the radical:

$$v^2 = 5000 + 24x$$

This linearizes the right side and makes the calculus incredibly straightforward.

### Final Calculation

Now, let's differentiate both sides of our squared equation with respect to $x$. Applying the power rule and the chain rule on the left side gives:

$$\frac{d}{dx}(v^2) = 2v \frac{dv}{dx}$$

On the right side, the derivative of the constant $5000$ is $0$, and the derivative of $24x$ is simply $24$. Equating them:

$$2v \frac{dv}{dx} = 24$$

Dividing both sides by $2$, we isolate our master term:

$$v \frac{dv}{dx} = 12$$

Since we already established that $a = v \frac{dv}{dx}$, we arrive directly at our final answer:

**$a = 12 	ext{ m/s}^2$**

### The Elegant Shortcut

There is an even faster way to solve this by recognizing the physical meaning of the equation. If we look at the third equation of motion for constant acceleration:

$$v^2 = u^2 + 2ax$$

And compare it directly to our squared equation:

$$v^2 = 5000 + 24x$$

By matching the coefficients of $x$, we immediately see that $2a = 24$, which gives $a = 12 	ext{ m/s}^2$. This confirms that the acceleration is indeed constant and provides a beautiful, calculus-free verification of our result!

Animated Solution for Physics - Kinematics: If the velocity of a body related to displacement $x$ is given by $v = 5000 + 24 x$ m/s, then the acceleration of the body is ...... m/s $^{2}$ .

Enter Numerical Value:

Visualized Solution

$v (x) Relationship$

$Acceleration Formula$

$Squaring the Equation$

$Differentiating w.r.t x$

$Calculating Acceleration$

$Kinematic Implications$

The Sigma Insight: Motion in a Straight Line

Analyzing the Setup

The Master Equation

The Smart Algebraic Move

Final Calculation

The Elegant Shortcut

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v(x) Relationship

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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Differentiating w.r.t x

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

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