Solution Question 26
Logic: Cars A and B move along the same straight road (let’s say X-axis). Car C moves on a different road. For C to be equidistant from A and B at all times ($r_{CA} = r_{CB}$), the projection of C’s position on the A-B line must always be the midpoint of A and B.
Therefore, the component of velocity of C along the direction of A and B ($v_{Cx}$) must be equal to the velocity of the midpoint of A and B.
$$ v_{mid} = \frac{v_A + v_B}{2} $$
$$ v_{Cx} = \frac{25 + 27}{2} = 26 \text{ m/s} $$
Calculation:
We know $|\vec{v}_C| = 30$ m/s. We found $v_{Cx} = 26$ m/s.
We can find the perpendicular component $v_{Cy}$:
$$ v_{Cy} = \sqrt{v_C^2 – v_{Cx}^2} = \sqrt{30^2 – 26^2} $$
$$ v_{Cy} = \sqrt{900 – 676} = \sqrt{224} \approx 14.96 \text{ m/s} $$
The question asks for relative velocity moduli: $|\vec{v}_{C/A}|$ and $|\vec{v}_{C/B}|$.
$\vec{v}_{C/A} = (v_{Cx} – v_A) \hat{i} + v_{Cy} \hat{j}$
$\vec{v}_{C/A} = (26 – 25) \hat{i} + \sqrt{224} \hat{j} = 1 \hat{i} + \sqrt{224} \hat{j}$
$|\vec{v}_{C/A}| = \sqrt{1^2 + (\sqrt{224})^2} = \sqrt{1 + 224} = \sqrt{225} = 15 \text{ m/s}$.
Similarly for B:
$\vec{v}_{C/B} = (26 – 27) \hat{i} + \sqrt{224} \hat{j} = -1 \hat{i} + \sqrt{224} \hat{j}$
$|\vec{v}_{C/B}| = \sqrt{(-1)^2 + 224} = \sqrt{225} = 15 \text{ m/s}$.
Answer: Relative velocity moduli are 15 m/s.
Logic: Cars A and B move along the same straight road (let’s say X-axis). Car C moves on a different road. For C to be equidistant from A and B at all times ($r_{CA} = r_{CB}$), the projection of C’s position on the A-B line must always be the midpoint of A and B.
Therefore, the component of velocity of C along the direction of A and B ($v_{Cx}$) must be equal to the velocity of the midpoint of A and B.
$$ v_{mid} = \frac{v_A + v_B}{2} $$ $$ v_{Cx} = \frac{25 + 27}{2} = 26 \text{ m/s} $$Calculation:
We know $|\vec{v}_C| = 30$ m/s. We found $v_{Cx} = 26$ m/s.
We can find the perpendicular component $v_{Cy}$:
$$ v_{Cy} = \sqrt{v_C^2 – v_{Cx}^2} = \sqrt{30^2 – 26^2} $$
$$ v_{Cy} = \sqrt{900 – 676} = \sqrt{224} \approx 14.96 \text{ m/s} $$
The question asks for relative velocity moduli: $|\vec{v}_{C/A}|$ and $|\vec{v}_{C/B}|$.
$\vec{v}_{C/A} = (v_{Cx} – v_A) \hat{i} + v_{Cy} \hat{j}$
$\vec{v}_{C/A} = (26 – 25) \hat{i} + \sqrt{224} \hat{j} = 1 \hat{i} + \sqrt{224} \hat{j}$
$|\vec{v}_{C/A}| = \sqrt{1^2 + (\sqrt{224})^2} = \sqrt{1 + 224} = \sqrt{225} = 15 \text{ m/s}$.
Similarly for B:
$\vec{v}_{C/B} = (26 – 27) \hat{i} + \sqrt{224} \hat{j} = -1 \hat{i} + \sqrt{224} \hat{j}$
$|\vec{v}_{C/B}| = \sqrt{(-1)^2 + 224} = \sqrt{225} = 15 \text{ m/s}$.