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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 3 / Problem 44P

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Solved: CALC The position of a dragonfly that is flying

Chapter 3, Problem 44P

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QUESTION:

CALC The position of a dragonfly that is flying parallel to the ground is given as a function of time by \(\vec{r}=\left[2.90 \mathrm{~m}+\left(0.0900 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\right] \hat{\imath}-\left(0.0150 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} \hat{\jmath}\) (a) At what value of 𝑡 does the velocity vector of the insect make an angle of 30.0° clockwise from the +𝑥-axis? (b) At the time calculated in part (a), what are the magnitude and direction of the acceleration vector of the insect?

Equation Transcription:

$\vec{r}$ $=\left[{2.90\ m\ +\ (0.0900\ m/{s}^{2}{)t}^{2}}\right]$ $î$ $-(0.0150\ m/{s}^{3}){t}^{3}$ $ĵ$

Text Transcription:

r =2.90 m + (0.0900 m/s2)t2î-(0.0150 m/s3)t3ĵ

Questions & Answers

QUESTION:

CALC The position of a dragonfly that is flying parallel to the ground is given as a function of time by \(\vec{r}=\left[2.90 \mathrm{~m}+\left(0.0900 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\right] \hat{\imath}-\left(0.0150 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} \hat{\jmath}\) (a) At what value of 𝑡 does the velocity vector of the insect make an angle of 30.0° clockwise from the +𝑥-axis? (b) At the time calculated in part (a), what are the magnitude and direction of the acceleration vector of the insect?

Equation Transcription:

$\vec{r}$ $=\left[{2.90\ m\ +\ (0.0900\ m/{s}^{2}{)t}^{2}}\right]$ $î$ $-(0.0150\ m/{s}^{3}){t}^{3}$ $ĵ$

Text Transcription:

r =2.90 m + (0.0900 m/s2)t2î-(0.0150 m/s3)t3ĵ

ANSWER:

Solution 44P

Step 1:

The position is given as,

Here the 2nd term in the $\widehat{i}$ is the acceleration.

$\ \ \ \ \ a=0.0900\ m/{s}^{2}\times{}{t}^{2}$

$\Rightarrow{}\frac{dv}{dt}=0.0900\ m/{s}^{2}\times{}{t}^{2}$

$\Rightarrow{}v=\int\limits_{\ }^{\ }0.0900\ {t}^{2}dt=0.0900\times{}\frac{{t}^{3}}{3}$

$\Rightarrow{}x=\int\limits_{\ }^{\ }vdt=\int\limits_{\ }^{\ }0.0900\times{}\frac{{t}^{3}}{3}=0.0900\times{}\frac{{t}^{4}}{12}=2.90+0.0075{\ t}^{4}$

The $\widehat{j}$ term has the dimensions of derivative of acceleration. So, we need to integrate it thrice to get the position as function of time.

$y=\int\limits_{\ }^{\ }\int\limits_{\ }^{\ }\int\limits_{\ }^{\ }0.0150\ {t}^{3}dt=0.000125\ {t}^{6}$ .

So, the position vector will be,

$\overline{r}\ =2.90+0.0075{\ t}^{4}\ \widehat{i}\ +0.000125\ {t}^{6}\ \widehat{j}$ -----------------(1)

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