$\begin{array}{l}
\overrightarrow{r_1}=2 t i-t^2 j+\left(3 t^2-4 t\right) k \\
\overrightarrow{r_2}=\left(5 t^2-12 t+4\right) i+t^3 j-3 t k
\end{array}$
Express in vector form, the radiative Velocity of the 2nd particle with respect to the 1st particle at [tex]$2 s$[/tex].
Find Velocity Vector of
[tex]$\vec{r}_1=2$[/tex]
Answer (2)
Find the velocity vector of particle 1 by differentiating its position vector: v 1 = 2 i ^ − 2 t j ^ + ( 6 t − 4 ) k ^ .
Find the velocity vector of particle 2 by differentiating its position vector: v 2 = ( 10 t − 12 ) i ^ + 3 t 2 j ^ − 3 k ^ .
Evaluate the velocity vectors at t = 2 : v 1 ( 2 ) = 2 i ^ − 4 j ^ + 8 k ^ and v 2 ( 2 ) = 8 i ^ + 12 j ^ − 3 k ^ .
Calculate the relative velocity: v 21 = v 2 − v 1 = 6 i ^ + 16 j ^ − 11 k ^ .
Explanation
Problem Analysis We are given the position vectors of two particles, r 1 and r 2 , as functions of time t . We need to find the relative velocity of the second particle with respect to the first particle at t = 2 seconds. This involves finding the velocity vectors of each particle by differentiating their position vectors with respect to time, evaluating these velocity vectors at t = 2 , and then subtracting the velocity vector of the first particle from the velocity vector of the second particle to find the relative velocity.
Velocity Vector of Particle 1 First, we find the velocity vector of particle 1 by differentiating its position vector r 1 with respect to time t :
r 1 = 2 t i ^ − t 2 j ^ + ( 3 t 2 − 4 t ) k ^ v 1 = d t d r 1 = d t d ( 2 t i ^ − t 2 j ^ + ( 3 t 2 − 4 t ) k ^ ) = 2 i ^ − 2 t j ^ + ( 6 t − 4 ) k ^
Velocity Vector of Particle 2 Next, we find the velocity vector of particle 2 by differentiating its position vector r 2 with respect to time t :
r 2 = ( 5 t 2 − 12 t + 4 ) i ^ + t 3 j ^ − 3 t k ^ v 2 = d t d r 2 = d t d (( 5 t 2 − 12 t + 4 ) i ^ + t 3 j ^ − 3 t k ^ ) = ( 10 t − 12 ) i ^ + 3 t 2 j ^ − 3 k ^
Evaluating Velocity Vectors at t=2 Now, we evaluate the velocity vectors v 1 and v 2 at t = 2 seconds: v 1 ( t = 2 ) = 2 i ^ − 2 ( 2 ) j ^ + ( 6 ( 2 ) − 4 ) k ^ = 2 i ^ − 4 j ^ + 8 k ^ v 2 ( t = 2 ) = ( 10 ( 2 ) − 12 ) i ^ + 3 ( 2 ) 2 j ^ − 3 k ^ = 8 i ^ + 12 j ^ − 3 k ^
Relative Velocity Calculation To find the relative velocity of particle 2 with respect to particle 1, we subtract the velocity vector of particle 1 from the velocity vector of particle 2: v 21 = v 2 − v 1 = ( 8 i ^ + 12 j ^ − 3 k ^ ) − ( 2 i ^ − 4 j ^ + 8 k ^ ) = ( 8 − 2 ) i ^ + ( 12 − ( − 4 )) j ^ + ( − 3 − 8 ) k ^ = 6 i ^ + 16 j ^ − 11 k ^
Final Answer Therefore, the relative velocity of the second particle with respect to the first particle at t = 2 seconds is: v 21 = 6 i ^ + 16 j ^ − 11 k ^
Examples
Understanding relative motion is crucial in many real-world applications. For example, in air traffic control, it's essential to know the relative velocities of aircraft to prevent collisions. Similarly, in robotics, when a robot arm is moving to grasp an object on a moving conveyor belt, the robot's control system needs to calculate the relative velocity between the arm and the object to ensure a successful grasp. These calculations rely on the principles of vector calculus and differentiation, just as we used in this problem.
To find the relative velocity of the second particle with respect to the first at t = 2 seconds, we first compute the velocity vectors for both particles and then subtract the first particle's velocity from the second. The result is v 21 = 6 i ^ + 16 j ^ − 11 k ^ .
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