Solve the problem.
-At time t = 0 a particle is located at the point (4, -3, 2). It travels in a straight line to the point (6, -2, 1), has speed 2 at (4, -3, 2) and constant acceleration 6i - j - k. Find an equation for the position vector r(t) of the particle at time t.
A) $r ( t ) = \left( 3 t ^ { 2 } + \frac { 8 } { \sqrt { 29 } } t + 4 \right) i - \left( \frac { 1 } { 2 } t ^ { 2 } + \frac { 6 } { \sqrt { 29 } } t + 3 \right) j - \left( \frac { 1 } { 2 } t ^ { 2 } - \frac { 4 } { \sqrt { 29 } } t - 2 \right) k$
B) $r ( t ) = \left( 3 t ^ { 2 } + \frac { 2 } { \sqrt { 6 } } t + 4 \right) i - \left( \frac { 1 } { 2 } t ^ { 2 } - \frac { 1 } { \sqrt { 6 } } t + 3 \right) j - \left( \frac { 1 } { 2 } t ^ { 2 } + \frac { 1 } { \sqrt { 6 } } t - 2 \right) k$
C) $\mathbf { r } ( \mathrm { t } ) = \left( 3 \mathrm { t } ^ { 2 } + \frac { 4 } { \sqrt { 6 } } \mathrm { t } + 4 \right) \mathrm { i } - \left( \frac { 1 } { 2 } \mathrm { t } ^ { 2 } - \frac { 2 } { \sqrt { 6 } } \mathrm { t } + 3 \right) \mathrm { j } - \left( \frac { 1 } { 2 } \mathrm { t } ^ { 2 } + \frac { 2 } { \sqrt { 6 } } \mathrm { t } - 2 \right) \mathrm { k }$
D) $\mathbf { r } ( \mathrm { t } ) = \left\{ 6 \mathrm { t } ^ { 2 } + \frac { 4 } { \sqrt { 6 } } \mathrm { t } + 4 \right) \mathrm { i } - \left( \mathrm { t } ^ { 2 } - \frac { 2 } { \sqrt { 6 } } \mathrm { t } + 3 \right) \mathrm { j } - \left( \mathrm { t } ^ { 2 } + \frac { 2 } { \sqrt { 6 } } \mathrm { t } - 2 \right) \mathbf { k }$

Question

Solve the problem.
-At time t = 0 a particle is located at the point (4, -3, 2). It travels in a straight line to the point (6, -2, 1), has speed 2 at (4, -3, 2) and constant acceleration 6i - j - k. Find an equation for the position vector r(t) of the particle at time t. 
A) $r ( t ) = \left( 3 t ^ { 2 } + \frac { 8 } { \sqrt { 29 } } t + 4 \right) i - \left( \frac { 1 } { 2 } t ^ { 2 } + \frac { 6 } { \sqrt { 29 } } t + 3 \right) j - \left( \frac { 1 } { 2 } t ^ { 2 } - \frac { 4 } { \sqrt { 29 } } t - 2 \right) k$
B) $r ( t ) = \left( 3 t ^ { 2 } + \frac { 2 } { \sqrt { 6 } } t + 4 \right) i - \left( \frac { 1 } { 2 } t ^ { 2 } - \frac { 1 } { \sqrt { 6 } } t + 3 \right) j - \left( \frac { 1 } { 2 } t ^ { 2 } + \frac { 1 } { \sqrt { 6 } } t - 2 \right) k$
C) $\mathbf { r } ( \mathrm { t } ) = \left( 3 \mathrm { t } ^ { 2 } + \frac { 4 } { \sqrt { 6 } } \mathrm { t } + 4 \right) \mathrm { i } - \left( \frac { 1 } { 2 } \mathrm { t } ^ { 2 } - \frac { 2 } { \sqrt { 6 } } \mathrm { t } + 3 \right) \mathrm { j } - \left( \frac { 1 } { 2 } \mathrm { t } ^ { 2 } + \frac { 2 } { \sqrt { 6 } } \mathrm { t } - 2 \right) \mathrm { k }$
D) $\mathbf { r } ( \mathrm { t } ) = \left\{ 6 \mathrm { t } ^ { 2 } + \frac { 4 } { \sqrt { 6 } } \mathrm { t } + 4 \right) \mathrm { i } - \left( \mathrm { t } ^ { 2 } - \frac { 2 } { \sqrt { 6 } } \mathrm { t } + 3 \right) \mathrm { j } - \left( \mathrm { t } ^ { 2 } + \frac { 2 } { \sqrt { 6 } } \mathrm { t } - 2 \right) \mathbf { k }$

Quizplus · Accepted Answer

The position vector $r(t)$ can be found by integrating the acceleration vector twice, given the initial velocity and position. The acceleration vector is $6i - j - k$, and the initial velocity can be found using the given speed and direction of motion. The direction from $(4, -3, 2)$ to $(6, -2, 1)$ is parallel to the vector $(6-4, -2+3, 1-2) = (2, 1, -1)$. The initial speed is 2, so the initial velocity vector $v(0)$ is proportional to $(2, 1, -1)$ and can be found by normalizing this direction vector and multiplying by the speed. The correct choice involves integrating the acceleration to get the velocity, then integrating again for position, and adjusting constants to match the initial conditions. Choice C correctly accounts for the acceleration and initial conditions.

Solve the Problem B) C) D)