University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 3 - Motion in Two or Three Dimensions - Problems - Exercises - Page 93: 3.3

Answer

a.) $v_{avg}=7.1cm/s$, $\theta=45^{\circ}$ b.) $v=5cm/s$, $\theta=90^{\circ}$ - $v_{avg}=7.1cm/s$, $\theta=45^{\circ}$ - $v_{avg}=11.1cm/s$, $\theta=27^{\circ}$ c.) (graph)

Work Step by Step

a.) In obtaining the average velocity using the given position vector, use the formula $$\vec{v_{avg}}=\frac{\vec{r}_{2}-\vec{r}_{1}}{t_{2}-t_{1}}$$ where $\vec{r}_2$ is the position vector at $t=2.0 s$ and $\vec{r}_1$ is the position vector at $t=0 s$. Substituting the change in time into the position vector, $$\vec{r}_{2}=[4.0cm+(2.5cm/s^{2})(2.0s)^{2}]\hat{i}+5.0cm/s(2.0s)\hat{j} =[4.0cm+10cm]\hat{i}+10cm\hat{j}=14cm\hat{i}+10cm\hat{j}$$ $$\vec{r}_{1}=[4.0cm+(2.5cm/s^{2})(0s)^{2}]\hat{i}+5.0cm/s(0s)\hat{j} =[4.0cm+0cm]\hat{i}+0cm\hat{j}=4cm\hat{i}$$ And then substituting the position vectors into the formula for average velocity, $$\vec{v_{avg}}=\frac{[14cm\hat{i}+10cm\hat{j}]-4cm\hat{i}}{2.0s-0s} =\frac{10cm\hat{i}+10cm\hat{j}}{2.0s}=10cm\hat{i}+10cm\hat{j}$$ Where the magnitude can be calculated using the following $$v_{avg}=\sqrt{(10cm/s)^{2}+(10cm/s)^{2}}=7.1cm/s$$ and the direction using tangent, $$\theta=\arctan\frac{10}{10}=45^{\circ}$$ b.) In the case of instantaneous velocity, the velocity formula changes into $$\vec{v}=v_{x}\hat{i}+v_{y}\hat{j}$$ Since a velocity vector can be defined as the first derivative of the position vector, evaluating the first derivative of the position vector $$\vec{v}=5cm/s(t)\hat{i}+5cm/s\hat{j}$$ Then finally substituting the given times, at $t=0s$, $$\vec{v}=5cm/s(0s)\hat{i}+5cm/s\hat{j}=5cm/s\hat{j}$$ $$v=\sqrt{(5cm/s)^{2}}=5cm/s$$ Since the vector is positive and $\hat{j}$, the direction is $90^{\circ}$. at $t=1s$, $$\vec{v}=5cm/s\hat{i}+5cm/s\hat{j}$$ This is similar to the answer in a. where $v=7.1cm/s$ and the direction $\theta=45^{\circ}$. at $t=2s$, $$\vec{v}=5cm/s(2s)\hat{i}+5cm/s\hat{j}=10cm/s\hat{i}+5cm/s\hat{j}$$ $$v=\sqrt{(10cm/s)^{2}+(5cm/s)^{2}}=11.1cm/s$$ Determining the direction using tangent, $$\theta=\arctan\frac{5}{10}=27^{\circ}$$ c. Using an online graphing calculator, input the position vector and substitute time points,
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