Question

The velocity of a particle moving along the \(x\) -axis is given, for \(t>0\), by \(v_{x}=\left(50.0 t-2.0 t^{3}\right) \mathrm{m} / \mathrm{s},\) where \(t\) is in seconds. What is the acceleration of the particle when (after \(t=0\) ) it achieves its maximum displacement in the positive \(x\) -direction?

Short answer

Short Answer: The acceleration of the particle when it achieves its maximum displacement in the positive x-direction is -100 \(\mathrm{m/s^2}\) after t=0.

Step-by-step solution

Find the position function

To find the position function, integrate the velocity function with respect to time, \(t\). Before integrating, we must also consider the integration constant, which we will represent as \(C\). \(x(t) = \int v_x \,\mathrm{dt} = \int (50.0t - 2.0t^3) \,\mathrm{dt} = 25t^2 - \frac{1}{2}t^4 + C\) To find the constant of integration, \(C\), we need the initial condition, which is given as the maximum displacement occurs after \(t=0\). So, the particle is at the origin in the beginning. At \(t=0\), \(x(0) = 0\) Substitute into the position function: \(0 = 25(0)^2 - \frac{1}{2}(0)^4 + C \implies C = 0\) So, the position function is \(x(t) = 25t^2 - \frac{1}{2}t^4\).

Identify the time of maximum positive displacement

To find when the maximum positive displacement occurs, we will set the velocity function equal to zero and find the corresponding time values. \(v_x = 50.0t - 2.0t^3 = 0\) Factor out the common term \(t\): \(t(50 - 2t^2) = 0\) This gives us two possible values for \(t\): 0 and \(\pm\sqrt{\frac{50}{2}}=\pm 5\) Since we're looking for the maximum positive displacement after \(t=0\), we discard the negative root, and \(t = 5\,\mathrm{s}\) is the time of maximum positive displacement.

Calculate the acceleration at the time of maximum positive displacement

To find the acceleration function, differentiate the velocity function with respect to time. \(a_x = \frac{\mathrm{d}v_x}{\mathrm{dt}} =\frac{\mathrm{d}}{\mathrm{dt}}(50.0t - 2.0t^3) = 50 - 6t^2\) Now, evaluate the acceleration function at the time of the maximum positive displacement, \(t = 5\,\mathrm{s}\): \(a_x (5) = 50 - 6(5^2) = 50 - 150 = -100\,\mathrm{m/s^2}\) After t=0, the acceleration of the particle when it achieves its maximum displacement in the positive x-direction is -100 \(\mathrm{m/s^2}\).

Key concepts

Velocity-Time Relationship

Understanding the relationship between the velocity of an object and time is crucial in the study of particle kinematics. Velocity, which is the rate of change of the particle's position, can be represented as a function of time. Mathematically, this is expressed as \(v(t)\), indicating how velocity varies as time progresses.

For instance, in the given exercise, the velocity function is provided as \(v_{x} = (50.0 t - 2.0 t^{3}) \frac{m}{s}\). Analyzing this function, we can observe that the velocity depends on time in a non-linear way. The term involving \(t^{3}\) suggests that the velocity will change depending on the cube of time, indicative of a more complex motion than simple linear motion.

To understand how this velocity affects the particle's movement, one needs to consider both the direction and the magnitude. A positive velocity implies forward movement, while a negative value suggests reversal. The point where velocity is zero is often of particular interest as it can represent a change in direction or a momentary pause in motion, which, as seen in this case, can help determine the time of the maximum positive displacement. This point is crucial to find before calculating the position function through integration or the acceleration from the derivative of the velocity function.

Position Function Integration

Once the velocity-time function is known, the position of the particle as a function of time can be ascertained through integration. Physically, integrating velocity with respect to time gives us the displacement, which when plotted alongside time, provides what is known as the position-time function or simply the position function, usually denoted by \(x(t)\).

In the context of our exercise, integrating the velocity function \(v_{x} = (50.0 t - 2.0 t^{3}) \frac{m}{s}\) with respect to time yields the position function \(x(t) = 25t^2 - \frac{1}{2}t^4 + C\) where \(C\) represents the integration constant. This constant agrees with the initial condition of the particle’s motion, which here is assumed to be starting from rest at the origin, thus setting \(C\) to zero.

This integrated position function is key in understanding the trajectory of the particle throughout its motion. If a student struggles with the concept of integrating the velocity function to find the position, one should look at it as a way of accumulating the distance travelled over time, piece by piece, with each small interval of time providing a tiny segment of the overall trajectory.

Acceleration Calculation

Acceleration is fundamentally the rate of change of velocity with respect to time, and it communicates how quickly the velocity of a particle is changing. It can be calculated by differentiating the velocity function, notated as \(a(t)\) or simply \(a\) when time is implicit.

By differentiating the provided velocity function \(v_{x} = (50.0 t - 2.0 t^{3}) \frac{m}{s}\), one obtains the acceleration as \(a_x = \frac{dv_x}{dt} = 50 - 6t^2 \frac{m}{s^2}\). This indicates that the acceleration is not constant but depends on the square of time, denoting a non-uniformly accelerated motion.

When seeking the acceleration at the time of maximum displacement, as in the exercise, it is essential to previously ascertain this particular time. After finding the time (which is when the velocity function equals zero, implying no change in displacement at that moment), the acceleration can be evaluated to provide a crucial insight into the behavior of the particle at that specific phase of its motion. Any student facing difficulties with this concept can imagine acceleration as a snapshot of how swiftly the particle's speed is increasing or decreasing at a specific moment in time.