The position vector of a body changes with time as vec r(t)…

The position vector of a body changes with time as $\vec r(t) = A t^2\,\vec i + B t\,\vec j + C\,\vec k$, where $A = 5\ \mathrm{m/s^2}$, $B = 3\ \mathrm{m/s}$, $C = 2\ \mathrm{m}$. Find: (a) the velocity and acceleration vectors; (b) the speed at $t = 2\ \mathrm{s}$; (c) the magnitude of the acceleration.

Explanation of Method

Velocity and acceleration from a position vector.

Step 1: (a) Velocity and acceleration

Velocity is the time derivative of position, taken component by component (a constant component gives zero): \[\vec v(t) = \frac{d\vec r}{dt} = 2At\,\vec i + B\,\vec j = 10t\,\vec i + 3\,\vec j\ \ (\mathrm{m/s}).\]

Acceleration is the time derivative of velocity: \[\vec a = \frac{d\vec v}{dt} = 2A\,\vec i = 10\,\vec i\ \ (\mathrm{m/s^2}).\]

\[\vec v(t) = 10t\,\vec i + 3\,\vec j, \qquad \vec a = 10\,\vec i\]

Step 2: (b) Speed at 2 s

\[ \vec v(2) = 20\,\vec i + 3\,\vec j, \]

\[ |\vec v(2)| = \sqrt{20^2 + 3^2} = \sqrt{409}. \]

\[|\vec v(2\ \mathrm{s})| \approx 20.2\ \mathrm{m/s}\]

Step 3: (c) Magnitude of the acceleration

\[|\vec a| = 10\ \mathrm{m/s^2} \ \ (\text{constant, along } x)\]

$y\ (\mathrm{m})$ $x\ (\mathrm{m})$ $20$ $5$ $3$ $6$ $t=0:\ \vec v = 3\vec j$ $t=1\ \mathrm{s}:\ \vec v = 10\vec i + 3\vec j$ $t=2\ \mathrm{s}:\ |\vec v| \approx 20.2\ \mathrm{m/s}$

The path in the plane $z = 2\ \mathrm{m}$ (seen from above). The $y$-part of the velocity stays 3 m/s while the $x$-part grows, so the velocity arrows turn toward the $x$-axis and get longer. (Arrows show direction; lengths are not to scale.)

Result

\[\boxed{\,\vec v(t) = 10t\,\vec i + 3\,\vec j, \qquad \vec a = 10\,\vec i\,}\]

\[\boxed{\,|\vec v(2\ \mathrm{s})| \approx 20.2\ \mathrm{m/s}\,}\]

\[\boxed{\,|\vec a| = 10\ \mathrm{m/s^2} \ \ (\text{constant, along } x)\,}\]

Remark

What the motion looks like. The $z$-coordinate stays 2 m, so the body moves in the plane $z = 2\ \mathrm{m}$. There it moves uniformly along $y$ (3 m/s) and with constant acceleration along $x$. Eliminating $t = y/3$ gives $x = \tfrac59 y^2$: a parabola, just like a projectile turned on its side.