A particle starts from the origin at t=0 with a velocity of 8.0 ĵ m / s and moves in the x y pla

A particle starts from the origin at t=0 with a velocity of...

Key Concepts

Vector Addition for Resultant Quantities

When dealing with quantities like velocity, which have both magnitude and direction, the individual components can be combined using vector addition. The overall speed is found by taking the square root of the sum of the squares of its components. This approach is essential in determining the resultant velocity of an object moving in two dimensions.

Kinematic Equations

These are the equations that describe the motion of objects under constant acceleration. They relate displacement, initial velocity, acceleration, and time. By using these equations for each component independently, one can solve for quantities such as the distance traveled or the speed of an object at a certain time.

Decomposition of Motion into Orthogonal Components

In two-dimensional motion, acceleration and velocity vectors are typically broken down into mutually perpendicular components (such as x and y directions). This decomposition allows each component to be analyzed independently using scalar kinematic equations, which simplifies the analysis of complex trajectories.

Uniformly Accelerated Motion

This concept refers to motion in which an object moves under a constant acceleration. The kinematic equations can be applied directly, allowing the prediction of displacement, velocity, and time. These equations are fundamental for analyzing projectile motion or any motion where the acceleration does not change over time.

Transcript

00:01 Hi there, so for this problem we have a particle with the given conditions at t equals to zero seconds.

00:08 The velocity is 8 meters per second only in the y component and it also moved with a constant acceleration that is also given 4 meters per second square in the x component and 2 meters per second square in the y component.

00:28 Now what we need to determine for the first part of this problem, part a, is the y component of its position when the x component or x coordinate of this particle is 29 meters.

00:49 Now in order to do that we can use the vector notation or we can use the coordinates separately the x component and the white component.

01:07 Now we are going to use that so the x component is equal to the initial speed the the x component of the initial speed times the time plus one half of the acceleration times the times squared.

01:30 And for the y component, we have something similar.

01:34 We have that that is the initial component of the speed times the time plus one half of the acceleration times the times square.

01:48 Now for the first and for the x component, we will have that we don't have in here any x component so the x component of that initial velocity is just zero so we don't have the first term the first term is zero and now in here we'll have the x component of the acceleration and in the other one the y component but we and we indeed have an x component of the acceleration so we just substitute that in there and solve for the time and then when we obtain the time we can substitute it in the other one to obtain the y coordinate so that's what we are going to do we know that solving for t we will have that that is two times x over the x company of the acceleration the square root of this will give us the time, so we just substitute those values.

02:59 We know that the x -m value, the x coordinate, is 29 meters.

03:06 The x component of the acceleration is four meters per second square, so from this we obtained that the time is 3 .8 seconds.

03:26 So with this, we can substitute in the second equation, for the y component or the y coordinate and we obtain the following we need the initial component of the white component the y component of the initial speed which is 8 so we put that in here 8 meters per seconds times the times that we are obtained 3 .8 seconds plus 1 half of the white component of the vector acceleration, which in this case you can see that is 2.

04:07 So we have 2 meters per second squared and the time that we obtain to the square.

04:15 So putting all of this into the calculator, we obtain that the y coordinate of this particle is 45 meters.

04:24 So this is the solution for the first part of this problem.

04:28 Now for part b, what we need to obtain is the speed of this particle.

04:41 The speed is the magnitude of the velocity...

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