A particle leaves the origin with its initial velocity given by v⃗0=11 +14 m / s, undergoing co

A particle leaves the origin with its initial velocity given...

Key Concepts

Constant Acceleration Kinematics

This concept involves using the kinematic equations that describe the motion of an object undergoing constant acceleration. It states that the displacement is given by the initial position plus the product of initial velocity and time, plus one-half the product of acceleration and the square of time. These equations are essential for predicting the future position and velocity of a particle when the acceleration remains fixed.

Vector Decomposition

In two-dimensional motion, vectors such as displacement, velocity, and acceleration are decomposed into their x and y components. This separation allows the independent analysis of the motion along each axis, simplifying the application of kinematic equations. It is particularly useful when different forces or motions influence each component differently.

Parametric Equations of Motion

The motion of a particle in a plane can be represented using parametric equations, where each coordinate (x and y) is expressed as a function of time. This approach enables one to solve for specific events—like the particle crossing the y-axis by setting the x-component to zero—and to analyze the path and behavior of the particle over time.

Velocity Magnitude and Direction

The velocity of a particle is a vector that has both magnitude (speed) and direction. Its magnitude is computed as the square root of the sum of the squares of its components. The direction is determined by the angle the velocity vector makes with a reference axis, often calculated using inverse trigonometric functions. This concept is critical in understanding not just how fast an object is moving, but also the path along which it travels.

Transcript

00:01 So for this exercise, we're given particles initial velocity and its acceleration.

00:07 And we're asked a couple of things.

00:11 So the initial velocity and acceleration are here on the screen.

00:15 And the first thing we're asked is at what time does the particle cross the y -axis? so let's draw a possible particle trajectory here in red.

00:30 Let's say this is the particle's trajectory.

00:36 This is just a supposition.

00:38 It's just in order to better illustrate the problem.

00:44 Notice that the particle crosses the y -axis when the value of the x -axis is zero.

00:58 So what we're going to have to do is to calculate the particle's trajectory and see when x -equalue.

01:05 X equals 0 and for that point the in that point the particle will be crossing the y -axis.

01:13 So let's write the particle's trajectory as usual r equals r0 plus v0 t plus the acceleration t squared over 2.

01:32 Here we have a vector and from this general formal we can separate r into coordinates.

01:44 So r is the position of the particle, we're gonna separate it in an x and y coordinate.

01:50 Let's first write the x coordinate.

01:54 X equals x0, x0 is the initial position of the particle, plus v0 x, the x components of the initial velocity, times t, plus ax, the x, the x component, the x components of the acceleration t squared over two.

02:17 Let me just separate things here.

02:20 Okay, so, well, we know that the particle, the exercise tells us that the particle starts its trajectory at the origin.

02:33 So x0 equals 0.

02:37 And we have the values of v0x, it's 11, and a x it's minus 1 .2.

02:48 And we want to solve this equation for x equals 0.

02:54 So we can simply do it.

02:57 First, let's write here that x equals 11 t minus 0 .6 t squared.

03:11 And we're going to make it equal to 0.

03:16 And we're going to solve this equation.

03:18 So let's solve it for t different from 0.

03:21 And the solution is t equals 11 over 0 .6, which is equal to 18 .3 seconds.

03:35 So the particle crosses the y -axis at this time, 18 .3 seconds.

03:42 This is the answer to the first question.

03:48 Now, the second question asks us, what is the value of why, when the particle crosses the y -axis.

03:57 So we're going to have to write the same equation as before the equation for the position of the particle.

04:06 But instead of writing it for component x, we're going to write it for component y.

04:12 So we're going to have y equals y0 plus v0y times t plus a y times t squared over two.

04:26 And the particle starts at the origin.

04:29 So y0 is going to be 0.

04:32 V0y, we already have this value.

04:38 It's 14.

04:42 And the time when the particle crosses the y -axis is 18 .3.

04:48 So we're going to multiply by 18 .3...

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