Question

two particles move in the x-y plane, exhibiting these velocity vectors as a function of time: (vec{v_1}(t) = a t hat{i} - b hat{j}); (vec{v_2}(t) = c hat{i} + d hat{j}) with: (a = 2.8 \frac{m}{s^2}), (b = 18.8 \frac{m}{s}), (c = 8.2 \frac{m}{s}), (d = 7.5 \frac{m}{s}) at one moment in time (measured by the variable (t) expressed in the equations above), these two particles collide, and when they do, their directions of motion are at right angles to each other. find this moment in seconds. provide at least one decimal place

Explanation:

Step1: Dot product of velocities = 0

For vectors at right angles, their dot product is zero:
$$\vec{v}_1(t) \cdot \vec{v}_2(t) = 0$$
Substitute the velocity vectors:
$$(At\hat{i} - B\hat{j}) \cdot (C\hat{i} + D\hat{j}) = 0$$
Expand the dot product:
$$At \cdot C + (-B) \cdot D = 0$$

Step2: Solve equation for t

Rearrange to isolate $t$:
$$ACt - BD = 0$$
$$t = \frac{BD}{AC}$$
Substitute the given values $A=2.8\ \frac{m}{s^2}$, $B=18.8\ \frac{m}{s}$, $C=8.2\ \frac{m}{s}$, $D=7.5\ \frac{m}{s}$:
$$t = \frac{18.8 \times 7.5}{2.8 \times 8.2}$$
Calculate numerator and denominator:
$$t = \frac{141}{22.96}$$
$$t \approx 6.1$$

Answer:

$6.1$ seconds