The paths of two particles for t ≥0 are given by the position vectors 𝐫1(t)=⟨ t-3,(t-3)^2⟩

The paths of two particles for t ≥0 are given by the...

The paths of two particles for $t \geq 0$ are given by the position vectors $$ \begin{array}{l} \mathbf{r}_{1}(t)=\left\langle t-3,(t-3)^{2}\right\rangle \\ \mathbf{r}_{2}(t)=\left\langle\frac{3 t}{2}-4, \frac{3 t}{2}-2\right\rangle \end{array} $$ (a) Determine the exact time(s) at which the particles collide. (b) Find the direction of motion of each particle at the time(s) of collision.

The paths of two particles for $t \geq 0$ are given by the position vectors
$$
\begin{array}{l}
\mathbf{r}_{1}(t)=\left\langle t-3,(t-3)^{2}\right\rangle \\
\mathbf{r}_{2}(t)=\left\langle\frac{3 t}{2}-4, \frac{3 t}{2}-2\right\rangle
\end{array}
$$
(a) Determine the exact time(s) at which the particles collide.
(b) Find the direction of motion of each particle at the time(s) of collision.

Key Concepts

Parametric Equations

Parametric equations are a way to describe the position of a moving object as a set of equations that depend on a parameter, typically time. In problems involving motion in the plane or space, each coordinate is expressed as a function of time, which allows for a clear depiction of the trajectory and motion dynamics.

Collision Detection in Particle Motion

Collision detection involves determining the common point(s) at which two moving objects meet. This is typically done by equating their position vectors or corresponding components and solving for the time parameter. It is a fundamental concept for analyzing interactions in dynamics and kinematics.

Systems of Equations

When equating the components of two parametric vectors, a system of equations is formed. Solving these simultaneous equations is essential to determine the specific time(s) at which the particles are at the same position, thereby verifying whether and when collisions occur.

Differentiation of Vector Functions

Differentiating a vector function with respect to time yields the velocity vector, which represents both the speed and the direction of motion of the particle. This process is central to kinematics and describes how an object’s position changes over time.

Velocity Vector and Direction of Motion

The velocity vector, obtained by differentiating the position vector, indicates the direction in which the particle is moving at any given moment. Its components show the rate of change along each coordinate axis, and analyzing this vector at a particular time gives insight into the particle's instantaneous direction and behavior at that time.

Transcript

00:01 In this question, we have the position vector for the two particles.

00:04 The first one is r1 t is equal to t minus 3 comma t minus 3 to the power 2.

00:16 And the second one is r2t is equals to 3 t by 2 minus 4 comma 3 t by 2 minus 2 minus 2.

00:35 We are required to find the time at which the particles collide and the direction of motion of each particles at the time of collision.

00:47 So let's see how to solve this question.

00:52 We know that the particles collide when the position vectors are equal.

01:22 Therefore, let's equate the x components of both the position vectors.

01:31 So we can write t minus 3 is equal to 3 t by 2 minus 4.

01:43 Hence, t minus 3 t by 2 is equals to 3 minus 4 and when we further calculate this, we get t is equals to 2.

02:03 Hence we can say that the particles collide at t is equal to 2.

02:19 So this is a final answer for the part a and now let's move to part b.

02:28 Let's find the velocity vector for the first position vector.

02:34 So we can write v1 t equals to d by d t of r1t.

02:43 Now, now, substitute the value of first position vector.

02:48 So we will have v1 t is equals to d by d t of t minus 3 comma t minus 3 to the power 2.

03:05 And we can write it as d by d d t of t minus 3 comma, d by d t minus 3 to the power 2.

03:22 Now differentiate these terms, so we get d by dt of t minus 3 will be equals to 1, comma, d by d t minus 3 to the power 2 will be equals to 2 multiplied by t minus 3.

03:43 And now substitute t is equal to 2...

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