Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 14 - Calculus of Vector-Valued Functions - 14.5 Motion in 3-Space - Exercises - Page 745: 27

Answer

At $t=10$ s, the position of the mass is at ${\bf{r}}\left( {10} \right) = \left( {45, - 20} \right)$.

Work Step by Step

We have constant force ${\bf{F}} = \left( {5,2} \right)$ and $m=10$. By Newton's second law: ${\bf{F}} = m{\bf{a}}$. So, $\left( {5,2} \right) = 10{\bf{a}}$, ${\ \ \ }$ ${\bf{a}} = \left( {\frac{1}{2},\frac{1}{5}} \right)$ 1. Find the velocity vector We have ${\bf{v}}\left( t \right) = \smallint {\bf{a}}\left( t \right){\rm{d}}t = \smallint \left( {\frac{1}{2},\frac{1}{5}} \right){\rm{d}}t = \left( {\frac{1}{2}t,\frac{1}{5}t} \right) + {{\bf{c}}_0}$ The initial condition ${\bf{v}}\left( 0 \right) = \left( {2, - 3} \right)$ gives $\left( {2, - 3} \right) = \left( {0,0} \right) + {{\bf{c}}_0}$ ${{\bf{c}}_0} = \left( {2, - 3} \right)$ Thus, ${\bf{v}}\left( t \right) = \left( {\frac{1}{2}t,\frac{1}{5}t} \right) + \left( {2, - 3} \right)$ ${\bf{v}}\left( t \right) = \left( {\frac{1}{2}t + 2,\frac{1}{5}t - 3} \right)$ 2. Find the position vector We have ${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {\frac{1}{2}t + 2,\frac{1}{5}t - 3} \right){\rm{d}}t$ ${\bf{r}}\left( t \right) = \left( {\frac{1}{4}{t^2} + 2t,\frac{1}{{10}}{t^2} - 3t} \right) + {{\bf{c}}_1}$ The initial condition ${\bf{r}}\left( 0 \right) = \left( {0,0} \right)$ gives $\left( {0,0} \right) = \left( {0,0} \right) + {{\bf{c}}_1}$ ${{\bf{c}}_1} = \left( {0,0} \right)$ Thus, the position of the mass at $t$ seconds is ${\bf{r}}\left( t \right) = \left( {\frac{1}{4}{t^2} + 2t,\frac{1}{{10}}{t^2} - 3t} \right)$ At $t=10$ s, the position of the mass is at ${\bf{r}}\left( {10} \right) = \left( {45, - 20} \right)$.
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