Question

For a particle moving along \(x\)-axis, the acceleration a of the particle in terms of its \(x\)-coordinate \(x\) is given by \(a=-9 x\), where \(x\) is in meters and a is in \(\mathrm{m} / \mathrm{s}^{2}\). Take acceleration, velocity and displacement in positive \(x\)-direction as positive. The initial velocity of particle at \(x=0\) is \(u=+6 \mathrm{~m} / \mathrm{s}\). The velocity of particle at \(x=2 \mathrm{~m}\) will be : \((1)+6 \sqrt{2} \mathrm{~m} / \mathrm{s}\) (2) \(-6 \sqrt{2} \mathrm{~m} / \mathrm{s}\) (3) \(72 \mathrm{~m} / \mathrm{s}\) (4) 0

Short answer

The velocity of the particle at \(x = 2 \, \mathrm{m}\) is 0 \(\mathrm{m/s}\).

Step-by-step solution

Understand the given information

The given acceleration is in terms of the position coordinate: \[a = -9x\] The initial velocity at position \(x = 0\) is given as \(u = +6 \, \mathrm{m/s}\). We need to find the velocity of the particle at position \(x = 2 \, \mathrm{m}\).

Relate acceleration to velocity

Acceleration \(a\) can be expressed as the derivative of the velocity \(v\) with respect to time \(t\): \[a = \frac{dv}{dt}\] Using the chain rule, we can express this as \[a = \frac{d v}{d t} = \frac{d v}{d x} \cdot \frac{d x}{d t} = \frac{d v}{d x} \cdot v\] Therefore, \[a = v \frac{d v}{d x}\]

Substitute acceleration into the equation

Substitute the given acceleration \(a = -9x\) into the equation from Step 2: \[-9x = v \frac{d v}{d x}\]

Integrate both sides

Rearrange the equation and integrate both sides to relate velocity \(v\) to position \(x\): \[\int v \, dv = -9 \int x \, dx\] The integrals are: \[\frac{v^2}{2} = -9 \cdot \frac{x^2}{2} + C\] Simplify to \[v^2 = -9x^2 + C\]

Determine the constant of integration

Use the initial condition where the initial velocity \(u\) is known at position \(x = 0\): \[6^2 = C\] This simplifies to \[C = 36\] Thus, the equation becomes \[v^2 = -9x^2 + 36\]

Solve for the velocity at x = 2 meters

Substitute \(x = 2\) into the equation: \[v^2 = -9(2)^2 + 36\]\[v^2 = -36 + 36\]\[v^2 = 0\]Therefore, \[v = 0 \, \mathrm{m/s}\]

Key concepts

Newton's laws of motion

Newton's laws of motion are fundamental principles that describe the relationship between the motion of an object and the forces acting on it.
They form the foundation of classical mechanics. There are three main laws:

First Law: An object will remain at rest or in uniform motion unless acted upon by an external force. This is also known as the law of inertia.
Second Law: The acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass, which is mathematically expressed as:
\[F = ma\]
where F is the force, m is the mass, and a is the acceleration.
Third Law: For every action, there is an equal and opposite reaction. This means that forces always occur in pairs.

Understanding these laws helps in solving problems related to particle motion along the x-axis, especially when dealing with acceleration, forces, and velocity changes.

Differential equations in mechanics

Differential equations are essential tools in mechanics as they describe how physical quantities change over time. In the context of particle motion, differential equations often relate velocity, acceleration, and position. For example:

If the acceleration a of a particle is given by a function of position x, such as:
\[a = -9x\]
We can use a differential equation to link this acceleration to velocity v and position x.
The acceleration a is the derivative of velocity with respect to time:
\[a = \frac{dv}{dt}\]
Using the chain rule, we can convert this to:
\[a = \frac{dv}{dx} \times \frac{dx}{dt} = \frac{dv}{dx} \times v\]
resulting in:
\[a = v \frac{dv}{dx}\]
Substituting the given acceleration function:
\[-9x = v \frac{dv}{dx}\]
This is a first-order differential equation and solving it involves integrating both sides,
which allows us to find the relationship between velocity and position.

Kinematics in one dimension

Kinematics is the branch of mechanics that describes the motion of objects without considering the forces that cause the motion. In one-dimensional kinematics, we focus on the motion along a straight line.

Key quantities include:

• Displacement (x): The change in position of a particle along the x-axis.
• Velocity (v): The rate of change of displacement with respect to time, which can be written as:
  \[v = \frac{dx}{dt}\]
• Acceleration (a): The rate of change of velocity with respect to time, expressed as:
  \[a = \frac{dv}{dt}\]

In this exercise, the acceleration is given as a function of position:
\[a = -9x\]
We use this information to find how velocity changes with position. By integrating the resulting differential equation, we find the velocity function in terms of position. This approach shows how kinematics in one dimension can be tied to calculus and differential equations to predict the motion of particles based on given initial conditions.