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\(^2\).

Hint:

When you see a velocity-displacement relationship of the form \(v = \sqrt{A + Bx}\) or \(v^2 = A + Bx\), immediately think of the kinematic equation \(v^2 = u^2 + 2ax\). This allows you to find the acceleration by simple comparison (\(2a = B\)) without needing to perform differentiation.

Answer 1

Correct Answer: 12

Step 1: Understanding the Question:
We are given the velocity of a body as a function of its displacement. We need to find its acceleration.
Step 2: Key Formula or Approach:
Acceleration (\(a\)) is the rate of change of velocity. When velocity is given as a function of position (\(x\)), the most convenient formula for acceleration is: \[ a = v \frac{dv}{dx} \] Alternatively, one can use the kinematic equation \(v^2 = u^2 + 2ax\).
Step 3: Detailed Explanation:
Method 1: Using Differentiation
Given velocity: \(v(x) = \sqrt{5000 + 24x} = (5000 + 24x)^{1/2}\).
First, we need to find the derivative of \(v\) with respect to \(x\), \(\frac{dv}{dx}\). We use the chain rule. \[ \frac{dv}{dx} = \frac{d}{dx} (5000 + 24x)^{1/2} \] \[ \frac{dv}{dx} = \frac{1}{2}(5000 + 24x)^{(1/2) - 1} \cdot \frac{d}{dx}(5000 + 24x) \] \[ \frac{dv}{dx} = \frac{1}{2}(5000 + 24x)^{-1/2} \cdot (24) \] \[ \frac{dv}{dx} = 12 (5000 + 24x)^{-1/2} = \frac{12}{\sqrt{5000 + 24x}} \] Now, use the formula for acceleration: \[ a = v \frac{dv}{dx} = (\sqrt{5000 + 24x}) \left( \frac{12}{\sqrt{5000 + 24x}} \right) \] The terms \(\sqrt{5000 + 24x}\) cancel out. \[ a = 12 \text{ m/s}^2 \] Method 2: Using Kinematic Equation
The given relation is \(v = \sqrt{5000 + 24x}\).
Square both sides: \[ v^2 = 5000 + 24x \] This equation is in the form of the third kinematic equation of motion for constant acceleration: \[ v^2 = u^2 + 2as \] Here, \(s\) is the displacement \(x\). So, \(v^2 = u^2 + 2ax\).
By comparing \(v^2 = 5000 + 24x\) with \(v^2 = u^2 + 2ax\), we can see that: - The initial velocity squared, \(u^2 = 5000\). - The term with \(x\) is \(2ax = 24x\). From this, we get \(2a = 24\).
\[ a = \frac{24}{2} = 12 \text{ m/s}^2 \] This shows that the acceleration is constant.
Step 4: Final Answer:
The acceleration of the body is 12 m/s\(^2\).