Acceleration of a Particle at x = 5

Given:
The displacement of a moving particle is given by the equation x = √(v^2 + 1), where v is the velocity of the particle.

To Find:
The acceleration of the particle when x = 5.

Explanation:

To find the acceleration of the particle at a particular displacement, we need to differentiate the displacement equation with respect to time. This will give us the expression for velocity, and then we can differentiate the velocity equation with respect to time to find the expression for acceleration.

Let's start by differentiating the displacement equation:

x = √(v^2 + 1)

Differentiating both sides of the equation with respect to time (t), we get:

dx/dt = d(√(v^2 + 1))/dt

Using the chain rule, we can simplify the expression:

dx/dt = (1/2)(v^2 + 1)^(-1/2) * d(v^2 + 1)/dt

Next, we need to differentiate the velocity equation, which is the derivative of the displacement equation:

v = dx/dt

Differentiating both sides of the equation with respect to time, we get:

dv/dt = d^2x/dt^2

Now, let's substitute the expression for dx/dt from the previous step into the second derivative equation:

dv/dt = d^2x/dt^2 = (1/2)(v^2 + 1)^(-1/2) * d(v^2 + 1)/dt

To find the acceleration at a specific displacement x = 5, we need to substitute this value into the velocity equation and solve for acceleration:

v = dx/dt = d(√(v^2 + 1))/dt

Let's solve this equation to find the value of v at x = 5:

5 = √(v^2 + 1)

Squaring both sides of the equation, we get:

25 = v^2 + 1

v^2 = 24

Taking the square root of both sides, we get:

v = ±√24

Since velocity cannot be negative, we take the positive root:

v = √24

Now, we can substitute this value of v into the expression for acceleration:

dv/dt = (1/2)(v^2 + 1)^(-1/2) * d(v^2 + 1)/dt

At x = 5, v = √24, so we have:

dv/dt = (1/2)(24 + 1)^(-1/2) * d(24 + 1)/dt

Simplifying further:

dv/dt = (1/2)(25)^(-1/2) * d(25)/dt

dv/dt = (1/2)(25)^(-1/2) * 0

dv/dt = 0

Therefore, the acceleration of the particle at x = 5 is zero.