# Math Help - Vectors problem

1. ## Vectors problem

If anyone could explain how the following problem is done, I'd be really grateful!

Given the acceleration vector a(t), the initial position r(0) = r_0, and the initial velocity v(0)=v_0 of a particle moving in space, find the position vector r(t) at time t.

a(t) = 6ti - 5j + 12t^2 k
r_0 = 3i + 4j
v_0 = 4j - 5k

2. From the given we get
$v(t) = \left( {3t^2 + C_1 } \right)i + \left( { - 5t + C_2 } \right)j + \left( {4t^3 + C_3 } \right)k$
$r(t) = \left( {t^3 + C_1 t + D_1 } \right)i + \left( {\frac{{ - 5t^2 }}{2} + C_2 t + D_2 } \right)j + \left( {t^4 + C_3 t + D_3 } \right)k$

Now use the initial values to find the C & D constants.

3. Originally Posted by Plato
From the given we get
$v(t) = \left( {3t^2 + C_1 } \right)i + \left( { - 5t + C_2 } \right)j + \left( {4t^3 + C_3 } \right)k$
$r(t) = \left( {t^3 + C_1 t + D_1 } \right)i + \left( {\frac{{ - 5t^2 }}{2} + C_2 t + D_2 } \right)j + \left( {t^4 + C_3 t + D_3 } \right)k$

Now use the initial values to find the C & D constants.
Sorry, I'm confused - how did you come up with those?

4. Hello, faure72!

Given: . $\begin{array}{cccc}\text{the acceleration vector:} & {\bf a}(t) & = & 6t{\bf i} - 5{\bf j} + 12t^2{\bf k} \\
\text{the position vector:} & {\bf r}_0 & = & 3{\bf i} + 4{\bf j} \\
\text{and the velocity vector:} & {\bf v}_0 & = & 4{\bf j} - 5{\bf k}\end{array}$

and: . $\begin{array}{cc}\text{initial position:} & {\bf r}(0) \:=\:{\bf r}_0 \\ \text{initial velocity:} & {\bf v}(0)\:=\:{\bf v}_0\end{array}$ . of a particle moving in space,

find the position vector ${\bf r}(t)$ at time $t$.

Integrate: . ${\bf v}(t) \;=\;\int\left(6t{\bf i} - 5{\bf j} + 12t^2{\bf k}\right)\,dt \;=\;3t^2{\bf i} - 5t{\bf j} + 4t^3{\bf k} + {\bf C}$

Since ${\bf v}(0) = {\bf v}_0$, we have: . $0^2\!\cdot\!{\bf i} - 0\!\cdot\!{\bf j} + 0\!\cdot\!{\bf k} + C \;=\;{\bf v}_0\quad\Rightarrow\quad {\bf C} \;=\;{\bf v}_0$

. . Hence: . ${\bf v}(t)\;=\;3t^2{\bf i} - 5t{\bf j} + 4t^3{\bf k} + {\bf v}_0$

Integrate: . $r(t) \;=\;\int\left(3t^2{\bf i} - 5t{\bf j} + 4t^3{\bf k} + {\bf v}_0\right)\,dt \;=\;t^3{\bf i} - \frac{5}{2}t^2{\bf j} + t^4{\bf k} + {\bf v}_ot + {\bf C}$

Since ${\bf r}(0) = {\bf r}_0$, we have: . $0\!\cdot\!{\bf i} - 0\!\cdot\!{\bf j} + 0\!\cdot\!{\bf k} + {\bf v}_0\!\cdot0 + {\bf C} \;=\;{\bf r}_0\quad\Rightarrow\quad {\bf C} \,=\,{\bf r}_0$

. . Therefore: . $\boxed{{\bf r}(t) \;=\;t^3{\bf i} - \frac{5}{2}t^2{\bf j} + t^4{\bf k} + {\bf v}_0t + {\bf r}_0}$