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Chapter 3: Problem 37

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

Expert verified
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

01

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: \(x(10)=-0.45 (10)^2-6.5 (10)+25\) \(y(10)=0.35 (10)^2+8.3 (10)+34\)
02

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: \(\textit{direction} =\tan^{-1}(\frac{y(10)}{x(10)})\)
03

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: \(\frac{dx}{dt}(10) = -0.9(10)-6.5\) \(\frac{dy}{dt}(10) = 0.7(10)+8.3\)
04

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: \(A_x = -0.9 \thinspace \text{m/s}^2\) \(A_y = 0.7 \thinspace \text{m/s}^2\)

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