Question

An object's velocity is measured to be \(v_x(t) =\alpha - \beta{t}^2\), where \(\alpha\) = 4.00 m/s and \(\beta\) = 2.00 m/s\(^3\). At \(t =\) 0 the object is at \(x =\) 0. (a) Calculate the object's position and acceleration as functions of time. (b) What is the object's maximum \(positive\) displacement from the origin?

Short answer

Position: \(x(t) = 4.00t - \frac{2.00 t^3}{3}\), Acceleration: \(a(t) = -4.00t\). Maximum displacement is \(\frac{8\sqrt{2}}{3}\) meters.

Step-by-step solution

Identify Given Information

We are given the velocity function \(v_x(t) = \alpha - \beta t^2\) where \(\alpha = 4.00\text{ m/s}\) and \(\beta = 2.00\text{ m/s}^3\). At \(t = 0\), the initial position \(x = 0\).

Calculate Acceleration

Acceleration is the derivative of velocity with respect to time. Thus, we differentiate \(v_x(t)\) with respect to \(t\): \[ a(t) = \frac{dv_x}{dt} = \frac{d}{dt}(\alpha - \beta t^2) = -2\beta t \] Substitute \(\beta = 2.00\text{ m/s}^3\): \[ a(t) = -2(2.00)t = -4.00t \text{ m/s}^2 \].

Calculate Position Function

Position is the integral of velocity with respect to time. We integrate \(v_x(t)\) to find \(x(t)\):\[ x(t) = \int v_x(t) \, dt = \int (\alpha - \beta t^2) \, dt = \alpha t - \frac{\beta t^3}{3} + C \] Given that \(x = 0\) at \(t = 0\), solve for \(C\): \[ 0 = \alpha(0) - \frac{\beta(0)^3}{3} + C \Rightarrow C = 0 \].Thus, \[ x(t) = 4.00t - \frac{2.00 t^3}{3} \].

Determine Maximum Positive Displacement

Maximum positive displacement occurs when velocity changes from positive to negative, i.e., when the velocity \(v_x(t)\) is zero. Set \(0 = \alpha - \beta t^2 \): \[ 0 = 4.00 - 2.00 t^2 \Rightarrow t^2 = 2 \Rightarrow t = \sqrt{2} \].Using \(t = \sqrt{2}\), compute \(x(t)\): \[ x(\sqrt{2}) = 4.00(\sqrt{2}) - \frac{2.00(\sqrt{2})^3}{3} \] \[ = 4\sqrt{2} - \frac{4\sqrt{2}}{3} = \frac{12\sqrt{2} - 4\sqrt{2}}{3} = \frac{8\sqrt{2}}{3} \text{ m} \].

Key concepts

Velocity Function

In kinematics, the velocity function describes how the speed of an object changes over time. In this scenario, we are given the velocity function as \(v_x(t) = \alpha - \beta t^2\), where \(\alpha\) and \(\beta\) are constants. This function shows that the velocity depends on both constant and time-dependent terms.

• \(\alpha\), the first term, represents the initial velocity when time \(t\) is zero. In our example, \(\alpha = 4.00 \text{ m/s}\), indicating that the initial velocity of the object is 4.00 m/s.
• The \(\beta t^2\) term shows how the velocity changes with time. The constant \(\beta = 2.00 \text{ m/s}^3\) influences the rate of this change, which in this case is a quadratic function, causing the velocity to decrease over time.

This function decreases over time because the quadratic term is subtracted, which indicates the object's speed is slowing down as time progresses.

Acceleration Calculation

Acceleration is a measure of how the velocity of an object changes with time. It is calculated by differentiating the velocity function \(v_x(t)\) with respect to time \(t\). In mathematical terms, this means finding the derivative.

For our velocity function \(v_x(t) = \alpha - \beta t^2\), the acceleration \(a(t)\) is given by \(a(t) = \frac{dv_x}{dt} = -2\beta t\).

• By substituting \(\beta = 2.00\text{ m/s}^3\) into the derivative, we obtain \(a(t) = -4.00t \text{ m/s}^2\).
• This implies that acceleration is a linear function of time, where the object's speed is decreasing at a constant rate as time goes on.

This negative acceleration signifies that the object is decelerating, meaning it is gradually coming to a stop before possibly reversing direction.

Position Function

The position function tells us where an object is located relative to a reference point as time progresses. It is obtained by integrating the velocity function \(v_x(t)\) over time.

The integration of \(v_x(t) = \alpha - \beta t^2\) gives us the position function \[x(t) = \int (\alpha - \beta t^2) \; dt = \alpha t - \frac{\beta t^3}{3} + C\].

• The integration constant \(C\) is determined using the initial condition that the position \(x\) is 0 at time \(t = 0\). Hence, \(C = 0\)
• This results in the position function \(x(t) = 4.00t - \frac{2.00t^3}{3}\).

This equation tells us how the object's position evolves over time.
The linear term \(4.00t\) represents the initial straightforward proportional increase in position, while the \(-\frac{2.00t^3}{3}\) term accounts for the adjustment in position due to acceleration.

Maximum Displacement

Maximum displacement from the origin is the furthest distance the object reaches before starting to return. It occurs at the instant when velocity becomes zero because the object momentarily stops moving forward before changing direction.

To find when velocity equals zero, solve the equation \(v_x(t) = 0\).

• Setting \(0 = \alpha - \beta t^2\), and substituting \(\alpha = 4.00 \text{ m/s}\) and \(\beta = 2.00 \text{ m/s}^3\), we find \(t = \sqrt{2}\).
• Substitute this \(t\) value into the position function \(x(t)\) we found earlier to get \(x(\sqrt{2}) = \frac{8\sqrt{2}}{3} \text{ m}\).

This value represents the maximum positive displacement of the object and is an essential concept in kinematics for understanding motion limits.