Q.The position of a particle is given by $\vec{r} = 3.0t\,\hat{i} + 2.0t^{2}\,\hat{j} + 5.0\,\hat{k}$ where $t$ is in seconds and the coefficients have the proper units for $\vec{r}$ to be in metres.

Concept: Kinematics Vector Differentiation — velocity is the first time derivative of position, acceleration is the derivative of velocity.

(a) Differentiate $\vec{r}$ component-wise:

(b) At $t = 1.0\ \text{s}$:

Magnitude:

Direction (angle $\theta$ from $+x$-axis):

The problem gives the position vector as a function of time:

All coefficients are in SI units so that $\vec{r}$ comes out in metres. The $z$-component is constant — the particle never moves in the $z$-direction.

Why differentiate?

In kinematics, velocity is the rate of change of position, and acceleration is the rate of change of velocity. Since $\vec{r}(t)$ is given in Cartesian components, we can differentiate each component separately — the unit vectors $\hat{i}, \hat{j}, \hat{k}$ are fixed in direction, so they behave like constants.

(a) Finding $\vec{v}(t)$ and $\vec{a}(t)$

1. Differentiate $\vec{r}(t)$ component by component:

   • $x$-component: $\displaystyle \frac{d}{dt}(3.0t) = 3.0$
   • $y$-component: $\displaystyle \frac{d}{dt}(2.0t^{2}) = 4.0t$
   • $z$-component: $\displaystyle \frac{d}{dt}(5.0) = 0$

   So the velocity vector is

Notice the $z$-component is zero — the motion is confined to the $xy$-plane.

2. Differentiate $\vec{v}(t)$ to get acceleration:

   • $x$-component: $\displaystyle \frac{d}{dt}(3.0) = 0$
   • $y$-component: $\displaystyle \frac{d}{dt}(4.0t) = 4.0$
   • $z$-component: $0$

   Hence

The acceleration is constant, purely in the $+y$ direction, with magnitude $4.0\ \text{m/s}^2$.

(b) Magnitude and direction of $\vec{v}$ at $t = 1.0\ \text{s}$

3. Plug $t = 1.0\ \text{s}$ into $\vec{v}(t)$: …

Concept: Velocity and Acceleration as Limits, in Vector Form

Method: First-Principles Limit Definition (no differentiation rules quoted)

Rather than applying $\frac{d}{dt}(t^n)=nt^{n-1}$ to each component directly, this method constructs $\vec v(t)$ and $\vec a(t)$ from the formal limit definitions $\vec v = \lim_{\Delta t\to0}\frac{\Delta \vec r}{\Delta t}$ and $\vec a=\lim_{\Delta t\to0}\frac{\Delta \vec v}{\Delta t}$ -- showing explicitly where each term in the final answer comes from.

Steps -- finding $\vec v(t)$

1. Write the position at $t+\Delta t$:

2. Expand and subtract $\vec r(t)$:

3. Divide by $\Delta t$:

4. Take the limit $\Delta t\to0$ -- the term $2.0\Delta t\to0$, leaving

Steps -- finding $\vec a(t)$

5. Repeat the same limit process on $\vec v(t)$:

(Already independent of $\Delta t$ -- consistent with the acceleration being constant.)

Steps -- magnitude and direction at $t=1.0\ \text{s}$

6. Substitute $t=1.0$ into $\vec v(t)$:

7. Magnitude: …