Question

1. suppose the position vector for a particle is given as a function of time by $vec{mathbf{r}}(t) = x(t)hat{mathbf{i}} + y(t)hat{mathbf{j}}$, with $x(t) = at + b$ and $y(t) = ct^2 + d$, where $a=1.00 \text{m/s}$, $b=1.00 \text{m}$, $c=0.125 \text{m/s}^2$, and $d = 1.00 \text{m}$. (a) calculate the average velocity during the time interval from $t = 2.00 \text{s}$ to $t = 4.00 \text{s}$. (b) determine the velocity and the speed at $t = 2.00 \text{s}$.

Explanation:

Step1: Define position at t=2.00 s

Calculate $x(2)$ and $y(2)$:
$x(2) = (1.00)(2.00) + 1.00 = 3.00\ \text{m}$
$y(2) = 0.125(2.00)^2 + 1.00 = 1.50\ \text{m}$
$\vec{r}(2) = 3.00\hat{\textbf{i}} + 1.50\hat{\textbf{j}}$

Step2: Define position at t=4.00 s

Calculate $x(4)$ and $y(4)$:
$x(4) = (1.00)(4.00) + 1.00 = 5.00\ \text{m}$
$y(4) = 0.125(4.00)^2 + 1.00 = 3.00\ \text{m}$
$\vec{r}(4) = 5.00\hat{\textbf{i}} + 3.00\hat{\textbf{j}}$

Step3: Compute average velocity (part a)

Average velocity is $\frac{\Delta\vec{r}}{\Delta t}$:
$\Delta\vec{r} = \vec{r}(4)-\vec{r}(2) = (5.00-3.00)\hat{\textbf{i}} + (3.00-1.50)\hat{\textbf{j}} = 2.00\hat{\textbf{i}} + 1.50\hat{\textbf{j}}$
$\Delta t = 4.00 - 2.00 = 2.00\ \text{s}$
$\vec{v}_{\text{avg}} = \frac{2.00\hat{\textbf{i}} + 1.50\hat{\textbf{j}}}{2.00} = 1.00\hat{\textbf{i}} + 0.75\hat{\textbf{j}}$

Step4: Find instantaneous velocity formula

Velocity is derivative of position:
$\vec{v}(t) = \frac{dx}{dt}\hat{\textbf{i}} + \frac{dy}{dt}\hat{\textbf{j}} = a\hat{\textbf{i}} + 2ct\hat{\textbf{j}}$

Step5: Compute velocity at t=2.00 s (part b)

Substitute $t=2.00$ s:
$\vec{v}(2) = 1.00\hat{\textbf{i}} + 2(0.125)(2.00)\hat{\textbf{j}} = 1.00\hat{\textbf{i}} + 0.50\hat{\textbf{j}}$

Step6: Calculate speed at t=2.00 s (part b)

Speed is magnitude of velocity:
$v = \sqrt{(1.00)^2 + (0.50)^2} = \sqrt{1.00 + 0.25} = \sqrt{1.25} = \frac{\sqrt{5}}{2} \approx 1.12\ \text{m/s}$

Answer:

(a) $\vec{v}_{\text{avg}} = 1.00\hat{\textbf{i}} + 0.75\hat{\textbf{j}}\ \text{m/s}$
(b) Velocity: $\vec{v}(2) = 1.00\hat{\textbf{i}} + 0.50\hat{\textbf{j}}\ \text{m/s}$; Speed: $\approx 1.12\ \text{m/s}$