Question

The acceleration of a particle is given by \(a_x(t) =\) -2.00 m/s\(^2\) + 13.00 m/s\(^3)t\). (a) Find the initial velocity \(v_{0x}\) such that the particle will have the same \(x\)-coordinate at \(t =\) 4.00 s as it had at \(t =\) 0. (b) What will be the velocity at \(t =\) 4.00 s?

Short answer

(a) Initial velocity: \(-30.67 \; \text{m/s}\). (b) Velocity at \(t=4.00 \; \text{s}\): \(65.33 \; \text{m/s}\).

Step-by-step solution

Understand the Motion Equation

The given equation for acceleration, which is the second derivative of position with respect to time, is \( a_x(t) = -2.00 \; \text{m/s}^2 + 13.00 \; \text{m/s}^3 \cdot t \). Integrating this equation with respect to time will give us the velocity function, and integrating the velocity function will give the position as a function of time.

Find the Velocity Function

Integrate the acceleration to find the velocity: \(\int a_x(t) dt = \int (-2.00 + 13.00t) dt = -2.00t + 13.00 \cdot \frac{t^2}{2} + C_1\). This simplifies to \(v_x(t) = -2.00t + 6.50t^2 + v_{0x}\), where \(v_{0x}\) is the constant of integration representing initial velocity.

Find the Position Function

Integrate the velocity equation to find the position: \(\int v_x(t) dt = \int (-2.00t + 6.50t^2 + v_{0x}) dt = -2.00 \cdot \frac{t^2}{2} + 6.50 \cdot \frac{t^3}{3} + v_{0x}t + C_2\). Simplifying, we have \(x(t) = -t^2 + \frac{6.50t^3}{3} + v_{0x}t + x_0\), where \(x_0\) is the initial position.

Set Boundary Condition to Solve for Initial Velocity

Since the particle should return to the initial position at \(t = 4.00 \; \text{s}\), set \(x(4) = x_0\): \(-16 + \frac{6.50 \cdot 64}{3} + 4v_{0x} = x_0\). This leads to \(-16 + 138.67 + 4v_{0x} = 0\). Solving for \(v_{0x}\) gives \(v_{0x} = \frac{-122.67}{4} = -30.67 \; \text{m/s}\).

Calculate the Velocity at t = 4.00 s

Substitute \(t = 4.00\) and \(v_{0x} = -30.67 \; \text{m/s}\) into the velocity equation: \(v_x(4) = -2.00 \cdot 4 + 6.50 \cdot 16 - 30.67\). Simplifying, \(v_x(4) = -8 + 104 - 30.67 = 65.33 \; \text{m/s}\).

Key concepts

Acceleration

In kinematics, acceleration refers to the rate at which an object's velocity changes over time. Here, the acceleration of the particle is described by the function \( a_x(t) = -2.00 \, \text{m/s}^2 + 13.00 \, \text{m/s}^3 \cdot t \). This equation shows that the acceleration varies with time, meaning it's not constant. The first term, \(-2.00 \, \text{m/s}^2\), represents a constant deceleration, while the second term, \(13.00 \, \text{m/s}^3 \cdot t\), signifies an increasing acceleration component that depends linearly on time.

• A positive acceleration means speeding up, while a negative acceleration means slowing down.
• Possible interpretations in real-world problems include any situation where a force changes over time, such as a rocket with thrust variations.

Understanding how acceleration affects motion is key in predicting future motion or reconstructing past behavior when analyzing the object's velocity and position.

Velocity Function

The velocity of an object is its speed in a given direction. Integration of the acceleration function yields the velocity function, helping understand how velocity changes over time.When we integrate the given acceleration function \( a_x(t) = -2.00 + 13.00t \), we perform:

• First, integrate: \( \int (-2.00 + 13.00t) dt = -2.00t + 6.50t^2 + C_1 \)
• The integration process introduces an integration constant \( C_1 \), equated to the initial velocity \( v_{0x} \).

This result, \( v_x(t) = -2.00t + 6.50t^2 + v_{0x} \), gives the velocity as a function of time, showing how both the linear and quadratic functions of time impact velocity.

Integration

Integration is a fundamental concept in calculus, used to find quantities like the velocity and position from a given acceleration. Understanding the process of integration involves reversing differentiation.

• Definite integration calculates the net change across an interval, while indefinite integration finds a family of functions with an arbitrary constant.
• In solving this problem, first, integrate the acceleration to obtain the velocity, then integrate the velocity to find the position.
• The integration process also uncovers two constants of integration, corresponding to initial velocity and initial position, which must be assessed with boundary conditions.

Through integration, you effectively "sum up" how small changes accumulate over a particular period.

Initial Velocity

Initial velocity \( v_{0x} \) is the starting velocity of a particle before any forces alter its speed or direction. It represents the constant of integration after integrating the acceleration function.

• Setting a condition at \( t = 4 \) helps determine \( v_{0x} \) numerically.
• The equation \(-16 + \frac{6.50 \cdot 64}{3} + 4v_{0x} = 0\) implies using \( v_x \) at two different times to ensure the particle returns to its starting position.
• Solving the boundary condition yields \( v_{0x} = -30.67 \, \text{m/s} \), which signifies the velocity needed to return the particle to its original position.

Knowing the initial velocity gives insights into how other forces or components will influence an object's path over time.