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Chapter 11: Q.49 (page 872)

Use the given acceleration vectors $a (t) = v (t) = r (t)$ and initial conditions in Exercises to find the position function $r (t)$ .
localid="1650737775645" $a (t) = 2 t, 3 t^{2}, v (0) = 1, 鈭� 2, r (0) = 2, 鈭� 3$

Short Answer

Expert verified

$r (t) = (\frac{r^{2}}{3} + t + 2 \frac{t^{2}}{4} - 2 i - 3)$

$r (t) = (\frac{t^{3}}{3} + t + 2, \frac{t^{4}}{4} - 2 t - 3)$

Step by step solution

01

Given Information

Consider $a (t) = (2 t, 3 t^{2}), v (0) = (1, - 2), r (0) = (2, - 3)$

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

Since $a (t) = v^{t} (t) = r^{n} (t)$ , we have $v (t) = 鈭� a (t) dt and r (t) = 鈭� v (t) dt$

Now,

But the given initial position $v (0) = (1, - 2) = i - 2 j . . . . . . . . . . . . . . (2)$

Comparing (1) and (2) equations, $C_{1} = 1, C_{2} = - 2$

02

Calculation

so

$v (t) = (t^{2} + 1) i + (t^{3} - 2) j$

Now,

$r (t) = 鈭� v (t 0 d t = 鈭� ((t^{2} + 1) i + (t^{2} - 2) j) d t$

$= (\frac{t^{3}}{3} + t) i + (\frac{t^{4}}{4} - 2 t) j + c$

Where c is a vector constant

$= (\frac{t^{3}}{3} + t + c_{3}) i + (\frac{t^{4}}{4} - 2 t + c_{4}) j$

where $c_{3}$ and $c_{4}$ are scalars.

$r (0) = (\frac{0^{3}}{3} + 0 + c_{3}) i + (\frac{0^{4}}{4} - 2 (0) + c_{4}) j r (0) = c_{3} i + c_{4} j . . . . . . . . . . (3)$

But as per problem, $r (0) = (2, - 3) = 2 i - 3 j . . . . . . . . . . . (4)$

Comparing $(3) & (4)$ equations, $c_{3} = 2, c_{4} = - 3$

Thus

$r (t) = (\frac{t^{3}}{3} + t + 2) i + (\frac{t^{4}}{4} - 2 t - 3) j = (\frac{t^{3}}{3} + t + 2, \frac{t^{4}}{4} - 2 t - 3)$

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