Q.52 Use the given acceleration vecto... [FREE SOLUTION]

Chapter 11: Q.52 (page 872)

Use the given acceleration vectors $a (t) = v' (t) = r " (t)$ and initial conditions in Exercises to find the position function $r (t)$ .
$a (t) = <\sin 3 t, t^{2}, \cos 3 t>, v (0) = <0, 0, 0>, r (0) = <2, - 5, 3>$

Short Answer

Expert verified

$r (t) = (\frac{- \sin 3 t}{9} + \frac{t}{3} + 2) i + (\frac{t^{4}}{12} - 5) j + (\frac{- \cos 3 t}{9} + \frac{28}{9}) k r (t) = <\frac{- \sin 3 t}{9} + \frac{t}{3} + 2, \frac{t^{4}}{12} - 5, \frac{- \cos 3 t}{9} + \frac{28}{9}>$

Step by step solution

Given Information

Consider,

$a (t) = <\sin 3 t, t^{2}, \cos 3 t>, v (0) = <0, 0, 0> r (0) = <2, - 5, 3>$

The objective is to find the position function $r (t)$ .

Since $a (t) = v' (t) = r " (t)$ , we have $v (t) = 鈭� a (t) d t$ and $r (t) = 鈭� v (t) d t$

Now

$v (t) d t = 鈭� a (t) d t = 鈭� <\sin 3 t, t^{2}, \cos 3 t> d t = 鈭� (\sin 3 t i + t^{2} j + \cos 3 t k) d t = \frac{- \cos 3 t}{3} i + \frac{t^{2}}{3} j + \frac{\sin 3 t}{3} k + c$

Where c is a constant

$v (t) = (\frac{- \cos 3 t}{3} + c_{1}) i + (\frac{t^{3}}{3} + c_{2}) j + (\frac{\sin 3 t}{3} + c_{3}) k$

Where $c_{1}, c_{2}$ and $c_{3}$ are scalars

$v (0) = (\frac{- 1}{3} + c_{1}) i + c_{2} j + c_{3} k . . . . . . . . . . . . . . (1)$

Expression

But as per problem, $v (0) = (0, 0, 0) = 0 i + 0 j + 0 k . . . . . . . . . . . . (2)$

Comparing (1) and (2) equations

$\frac{- 1}{3} + c_{1} = 0, c_{2} = 0, c_{3} = 0 c_{1} = \frac{1}{3} s o v (t) = (\frac{- \cos 3 t + 1}{3}) i + \frac{t^{2}}{3} j + \frac{\sin 3 t}{3} k$

Now

$r (t) = 鈭� v (t) d t = 鈭� (\frac{- \cos 3 t + 1}{3}) i + \frac{t^{2}}{3} j + \frac{\sin 3 t}{3} k d t r (0) = c_{4} i + c_{5} j + (\frac{- \cos 0}{9} + c_{6}) k r (0) = c_{4} i + c_{5} j + (\frac{- 1}{9} + c_{6}) k . . . . . . . . . . . . (3) r (0) = (2, - 5, 3) = 2 i - 5 j + 3 k . . . . . . . . . . . . . (4)$

Comparing (3) and (4) equations,

$c_{4} = 2, c_{5} = - 5 \frac{- 1}{9} + c_{6} = 3$

$C_{6} = 3 + \frac{1}{9} C_{6} = \frac{28}{9}$

$r (t) = (\frac{- \sin 3 t}{9} + \frac{1}{3} + 2 \frac{t^{3}}{12} 5, \frac{- \cos 3 t}{9} + \frac{28}{9})$

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91影视

Use the given acceleration vectors $a (t) = v' (t) = r " (t)$ and initial conditions in Exercises to find the position function $r (t)$ .
$a (t) = <\sin 3 t, t^{2}, \cos 3 t>, v (0) = <0, 0, 0>, r (0) = <2, - 5, 3>$

Short Answer

Step by step solution

Given Information

Expression

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91影视

Use the given acceleration vectors a(t)=v'(t)=r"(t)and initial conditions in Exercises to find the position function r(t).a(t)=sin3t,t2,cos3t,v(0)=0,0,0,r(0)=2,-5,3

Short Answer

Step by step solution

Given Information

Expression

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Mechanics Maths

Decision Maths

Pure Maths

Calculus

Geometry

Study anywhere. Anytime. Across all devices.

Use the given acceleration vectors $a (t) = v' (t) = r " (t)$ and initial conditions in Exercises to find the position function $r (t)$ .
$a (t) = <\sin 3 t, t^{2}, \cos 3 t>, v (0) = <0, 0, 0>, r (0) = <2, - 5, 3>$