Q33: Speed of Particle C at $t = 5\text{s}$

At $t = 5$, the separation $AB = 100 – 10(5) = 50 \text{ m}$. The sides of the triangle are $AC=40, BC=30, AB=50$. Since $30^2 + 40^2 = 50^2$, the triangle is right-angled at C.

Let the origin be at the midpoint of AB. $A = (-25, 0)$, $v_A = (5, 0)$ $B = (25, 0)$, $v_B = (-5, 0)$ $C = (x, y)$. Using geometry for the 30-40-50 triangle: $x = 7, y = 24$. So $C(7, 24)$.

Since the distances $AC$ and $BC$ are constant, the relative velocity along the rod must be zero: 1. $\vec{r}_{CA} \cdot (\vec{v}_C – \vec{v}_A) = 0$ 2. $\vec{r}_{CB} \cdot (\vec{v}_C – \vec{v}_B) = 0$

Let $\vec{v}_C = (u, v)$. $\vec{r}_{CA} = C – A = (32, 24)$. $\vec{r}_{CB} = C – B = (-18, 24)$.

Equation 1: $(32, 24) \cdot (u-5, v) = 0 \Rightarrow 32(u-5) + 24v = 0 \Rightarrow 4u + 3v = 20$.
Equation 2: $(-18, 24) \cdot (u+5, v) = 0 \Rightarrow -18(u+5) + 24v = 0 \Rightarrow -3u + 4v = 15$.

Solving the system: Multiply (1) by 3 and (2) by 4: $12u + 9v = 60$ $-12u + 16v = 60$ Summing them: $25v = 120 \Rightarrow v = 4.8 \text{ m/s}$. Substitute back: $4u + 3(4.8) = 20 \Rightarrow 4u = 5.6 \Rightarrow u = 1.4 \text{ m/s}$.

Speed $|\vec{v}_C| = \sqrt{1.4^2 + 4.8^2} = \sqrt{1.96 + 23.04} = \sqrt{25} = 5 \text{ m/s}$.

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Q34: Modulus of Acceleration of C at $t = 5\text{s}$

Differentiating the velocity constraint $\vec{r} \cdot \vec{v}_{rel} = 0$ gives: $$ |\vec{v}_{rel}|^2 + \vec{r} \cdot \vec{a}_{rel} = 0 $$ Since particles A and B move with constant velocity, $\vec{a}_A = 0$ and $\vec{a}_B = 0$. Thus $\vec{a}_{rel} = \vec{a}_C$. $$ \vec{r} \cdot \vec{a}_C = -|\vec{v}_{rel}|^2 $$

Calculate relative speeds squared: $\vec{v}_{rel, A} = \vec{v}_C – \vec{v}_A = (1.4 – 5, 4.8) = (-3.6, 4.8)$. $|\vec{v}_{rel, A}|^2 = (-3.6)^2 + (4.8)^2 = 12.96 + 23.04 = 36$.
$\vec{v}_{rel, B} = \vec{v}_C – \vec{v}_B = (1.4 – (-5), 4.8) = (6.4, 4.8)$. $|\vec{v}_{rel, B}|^2 = (6.4)^2 + (4.8)^2 = 40.96 + 23.04 = 64$.

System of Equations for $\vec{a}_C = (a_x, a_y)$: 1. $\vec{r}_{CA} \cdot \vec{a}_C = -36 \Rightarrow 32a_x + 24a_y = -36 \Rightarrow 8a_x + 6a_y = -9$. 2. $\vec{r}_{CB} \cdot \vec{a}_C = -64 \Rightarrow -18a_x + 24a_y = -64 \Rightarrow -9a_x + 12a_y = -32$.

Solving: Multiply (1) by 2: $16a_x + 12a_y = -18$. Subtract (2) from modified (1): $(16 – (-9))a_x = -18 – (-32) \Rightarrow 25a_x = 14 \Rightarrow a_x = 0.56$. Substitute $a_x$ into (1): $8(0.56) + 6a_y = -9 \Rightarrow 4.48 + 6a_y = -9 \Rightarrow 6a_y = -13.48 \Rightarrow a_y \approx -2.25$.

Magnitude $|\vec{a}_C| = \sqrt{(0.56)^2 + (-2.25)^2} \approx \sqrt{0.31 + 5.06} = \sqrt{5.37} \approx 2.3 \text{ m/s}^2$.