Question

The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t equals 0.

Answer 1

Answer:

$\theta = n\pi/2, {\rm where~n~is~an~integer.}$

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

$\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k$

Similarly,

$\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k$

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

$\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)$

So,

$\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)$

At time t = 0, this equation becomes

$0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}$