Question

d) for ( 0 leq t leq 10 ), particles ( p ) and ( q ) move along the ( x ) axis. the position of particle ( p ) can be modeled by ( x_{p}(t) ) as shown in the figure above. the position of particle ( q ) is defined by ( x_{q}(t) ). selected values of ( x_{q}(t), v_{q}(t) ), and ( a_{q}(t) ) are given in the table above. at what time ( t ) are particles ( p ) and ( q ) moving towards each other?

Explanation:

Step1: Find the velocity of particle \(P\)

The velocity of particle \(P\), \(v_{P}(t)\) is the slope of the position - time graph \(x_{P}(t)\).
For \(0\leq t\leq3\), \(v_{P}(t)=\frac{0 - 3}{3-0}=- 1\)
For \(3\leq t\leq8\), \(v_{P}(t)=\frac{0.5+1}{8 - 3}=\frac{1.5}{5}=0.3\)
For \(8\leq t\leq10\), \(v_{P}(t)=\frac{-3 - 0.5}{10 - 8}=\frac{-3.5}{2}=-1.75\)

Step2: Analyze the condition for moving towards each other

Two particles are moving towards each other when \(v_{P}(t)\times v_{Q}(t)<0\) (opposite - direction velocities)

• At \(t = 1\): \(v_{P}(1)=-1\), \(v_{Q}(1) = 3\), and \(v_{P}(1)\times v_{Q}(1)=(-1)\times3=-3<0\)
• At \(t = 3\): \(v_{P}(3)=-1\), \(v_{Q}(3)=0\), and \(v_{P}(3)\times v_{Q}(3)=(-1)\times0 = 0\)
• At \(t = 7\): \(v_{P}(7)=0.3\), \(v_{Q}(7)=-2\), and \(v_{P}(7)\times v_{Q}(7)=(0.3)\times(-2)=-0.6<0\)
• At \(t = 9\): \(v_{P}(9)=-1.75\), \(v_{Q}(9)=3\), and \(v_{P}(9)\times v_{Q}(9)=(-1.75)\times3=-5.25<0\)

Answer:

\(t = 1\), \(t = 7\), \(t = 9\)