The vector r(t) is the position vector of a particle at time t. Find the angle between the velocity and the acceleration vectors at time t = 0.
-$$r ( t ) = \left( 3 t ^ { 2 } + 3 \right) i + \left( 4 t ^ { 3 } - 2 t \right) k$$
A) 0
B) $\pi$
C) $\frac { \pi } { 4 }$
D) $\frac { \pi } { 2 }$

Question

Quizplus · Accepted Answer

The velocity vector $v(t)$ is the derivative of the position vector $r(t)$ with respect to time $t$, and the acceleration vector $a(t)$ is the derivative of the velocity vector with respect to time. For the given position vector $r(t) = (3t^2 + 3)i + (4t^3 - 2t)k$, we find the velocity and acceleration vectors as follows:1. Velocity vector $v(t) = \frac{dr(t)}{dt} = (6t)i + (12t^2 - 2)k$.2. Acceleration vector $a(t) = \frac{dv(t)}{dt} = 6i + (24t)k$.At $t = 0$, the velocity vector $v(0) = 6i - 2k$ and the acceleration vector $a(0) = 6i$.The angle $\theta$ between the velocity and acceleration vectors can be found using the dot product formula: $$v \cdot a = |v||a|\cos(\theta)$$At $t = 0$, $v(0) \cdot a(0) = (6i - 2k) \cdot (6i) = 36 - 0 = 36$, and the magnitudes are $|v(0)| = \sqrt{6^2 + (-2)^2} = \sqrt{40}$ and $|a(0)| = \sqrt{6^2} = \sqrt{36}$.Since the dot product is positive and both vectors have non-zero magnitudes, but the acceleration vector has no component in the $k$ direction while the velocity does, it indicates they are perpendicular at $t = 0$. Thus, the angle between them is $\frac{\pi}{2}$ or 90 degrees.

The Vector R(t) Is the Position Vector of a Particle r(t)=(3t2+3)i+(4t3−2t)kr ( t ) = \left( 3 t ^ { 2 } + 3 \right) i + \left( 4 t ^ { 3 } - 2 t \right) kr(t)=(3t2+3)i+(4t3−2t)k

The Vector R(t) Is the Position Vector of a Particle $r ( t ) = \left( 3 t ^ { 2 } + 3 \right) i + \left( 4 t ^ { 3 } - 2 t \right) k$