Question

A rabbit runs in a garden such that the \(x\) - and \(y\) components of its displacement as function of times are given by \(x(t)=-0.45 t^{2}-6.5 t+25\) and \(y(t)=0.35 t^{2}+8.3 t+34 .\) (Both \(x\) and \(y\) are in meters and \(t\) is in seconds.) a) Calculate the rabbit's position (magnitude and direction) at \(t=10 \mathrm{~s}\) b) Calculate the rabbit's velocity at \(t=10 \mathrm{~s}\). c) Determine the acceleration vector at \(t=10 \mathrm{~s}\).

Short answer

Short Answer: At t=10 s, the rabbit's position vector has an x-component of \(x(10)=-0.45 (10)^2-6.5 (10)+25\) and a y-component of \(y(10)=0.35 (10)^2+8.3 (10)+34\). The velocity vector has an x-component of \(\frac{dx}{dt}(10) = -0.9(10)-6.5\) and a y-component of \(\frac{dy}{dt}(10) = 0.7(10)+8.3\). The acceleration vector has an x-component of \(A_x = -0.9 \thinspace \text{m/s}^2\) and a y-component of \(A_y = 0.7 \thinspace \text{m/s}^2\).

Step-by-step solution

Find the position vector at t=10 s

Plug in t=10 s into the expressions for x(t) and y(t) to find the components of the position vector: \(x(10)=-0.45 (10)^2-6.5 (10)+25\) \(y(10)=0.35 (10)^2+8.3 (10)+34\)

Calculate the magnitude and direction of the position vector

Calculate the magnitude of the position vector using the Pythagorean theorem: \(\textit{magnitude} =\sqrt{x(10)^2+y(10)^2}\) Then, calculate the direction of the position vector using the arctangent function: \(\textit{direction} =\tan^{-1}(\frac{y(10)}{x(10)})\)

Calculate the velocity vector at t=10 s

To determine the velocity vector, we need to calculate the derivative of the position vector with respect to time. For both x(t) and y(t) components, we have: \(\frac{dx}{dt}=-0.9t-6.5\) \(\frac{dy}{dt}=0.7t+8.3\) Now, plug in t=10 s into these expressions to find the components of the velocity vector: \(\frac{dx}{dt}(10) = -0.9(10)-6.5\) \(\frac{dy}{dt}(10) = 0.7(10)+8.3\)

Calculate the acceleration vector at t=10 s

To determine the acceleration vector, we need to calculate the derivative of the velocity vector with respect to time. For both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) components, we have: \(\frac{d^2x}{dt^2}=-0.9\) \(\frac{d^2y}{dt^2}=0.7\) Since the acceleration vector components do not depend on time, the acceleration vector at t=10 s will be the same: \(A_x = -0.9 \thinspace \text{m/s}^2\) \(A_y = 0.7 \thinspace \text{m/s}^2\)

Key concepts

Position Vector

The position vector is a fundamental concept in kinematics, used to describe the location of an object in space. In this exercise, the rabbit's position vector at time \( t = 10 \) seconds is calculated for both the \( x \) and \( y \) components. Given the equations \( x(t) = -0.45 t^2 - 6.5 t + 25 \) and \( y(t) = 0.35 t^2 + 8.3 t + 34 \), you plug in \( t = 10 \) to find these components. These components form a vector \( \mathbf{r}(t) \) pointing from the origin to the rabbit's position. The magnitude of this vector can be computed using the Pythagorean theorem, \( \sqrt{x(10)^2 + y(10)^2} \), which gives a sense of the rabbit's distance from the origin. Additionally, the direction relative to the x-axis can be found using \( \tan^{-1}(\frac{y(10)}{x(10)}) \), providing a full description of its position.

Velocity Vector

The velocity vector is key to understanding how quickly and in what direction the rabbit is moving. It is calculated by taking the first derivative of the position vector's components with respect to time. For the given functions, \( x(t) \) and \( y(t) \), the derivatives \( \frac{dx}{dt} = -0.9 t - 6.5 \) and \( \frac{dy}{dt} = 0.7 t + 8.3 \) represent the velocity's components. At \( t = 10 \) seconds, these derivatives give the rabbit's velocity in both directions. Evaluating these derivatives at \( t = 10 \), you determine the specific numerical values for the velocity components, furnishing a comprehensive view of the rabbit's speed and heading at that moment.

Acceleration Vector

Acceleration defines how the velocity of the rabbit changes over time. It's obtained as the derivative of the velocity vector with respect to time, or the second derivative of the position function. For the rabbit, the acceleration components, \( \frac{d^2x}{dt^2} = -0.9 \) and \( \frac{d^2y}{dt^2} = 0.7 \), remain constant since they do not depend on \( t \). This uniform acceleration indicates that the forces acting on the rabbit are consistent over time, giving fixed values for both acceleration components. These values describe how much the velocity's \( x \) and \( y \) components change each second.

Displacement as a Function of Time

Displacement provides the change in position of the rabbit over time. It is represented by the functions \( x(t) \) and \( y(t) \), mapping how the rabbit moves along the x and y axes. Displacement accounts for both magnitude and direction, thus giving a comprehensive description of the motion. As \( t \) varies, you see how the position vector changes, providing insights into the path traveled. By observing these functions, one can predict where the rabbit will be at any specific time, assuming the motion described continues uninterrupted.

Derivatives in Physics

Derivatives in physics allow us to relate various physical concepts like velocity and acceleration. These mathematical tools describe change; the first derivative of a position function gives velocity, showing the rate of changing displacement. Similarly, the second derivative gives acceleration, indicating the rate of change of velocity. These derivatives transform information from a pure description of position into dynamic insights about movement, permitting predictions. Understanding how to compute and interpret these derivatives equips one with the ability to analyze motion comprehensively, as showcased in this exercise where derivatives provide both velocity and acceleration at any given time.