Question

question 2

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 } \quad \vec { v } _ { 2 } ( t ) = c \hat { i } + d \hat { j } \\)

with:

\\( a = 4.8 \frac { m } { s ^ { 2 } }, b = 30.6 \frac { m } { s }, c = 7.8 \frac { m } { s }, d = 7.8 \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: Use the dot - product formula for perpendicular vectors

If two vectors \(\vec{v}_1=(A t)\hat{i}-B\hat{j}\) and \(\vec{v}_2 = C\hat{i}+D\hat{j}\) are perpendicular, then \(\vec{v}_1\cdot\vec{v}_2 = 0\).
The dot - product formula is \(\vec{v}_1\cdot\vec{v}_2=(A t)\times C+(-B)\times D\).
Substitute \(A = 4.8\frac{m}{s^{2}}\), \(B = 30.6\frac{m}{s}\), \(C = 7.8\frac{m}{s}\), \(D = 7.8\frac{m}{s}\) into the dot - product formula:
\((4.8t)\times7.8+(- 30.6)\times7.8 = 0\).

Step2: Simplify the equation

First, factor out \(7.8\): \(7.8(4.8t - 30.6)=0\).
Since \(7.8
eq0\), we can solve the equation \(4.8t-30.6 = 0\).
Add \(30.6\) to both sides: \(4.8t=30.6\).
Then divide both sides by \(4.8\): \(t=\frac{30.6}{4.8}\).

Step3: Calculate the value of \(t\)

\(t=\frac{30.6}{4.8}=\frac{306}{48}=\frac{51}{8}=6.375\approx6.4\)

Answer:

\(6.4\)