A spacecraft is traveling with a velocity of v0 x=5480 m / s...

A spacecraft is traveling with a velocity of $v_{0 x}=5480 {m} / {s}$ along the $+x$ direction. Two engines are turned on for a time of 842 s. One engine gives the spacecraft an acceleration in the $+x$ direction of $a_{x}=1.20 {m} / {s}^{2}$ while the other gives it an acceleration in the $+y$ direction of $a_{y}=8.40 {m} / {s}^{2}$ At the end of the firing, find (a) $v_{x}$ and $({b}) v_{y}$

Key Concepts

Vector Component Analysis

This principle involves breaking down a vector into its orthogonal components, which in the context of motion allows for separate analysis of movements along the x-axis and y-axis. By treating each component independently, one can accurately calculate the effects of forces or accelerations applied in different directions, ensuring a clear understanding of the overall trajectory.

Two-Dimensional Kinematics

This concept refers to analyzing motion in two independent directions, typically represented as the x and y axes. In such analyses, the motion along one axis does not influence the motion along the other, allowing us to separately compute changes in velocity and displacement for each component when provided with initial conditions and relevant acceleration values.

Constant Acceleration Equations

Under constant acceleration, the relationship between initial velocity, acceleration, and time is given by the equation v = v0 + at for each directional component. This equation is fundamental in computing the final velocity of an object, as it directly links the known initial motion, any applied acceleration, and the accumulated change over time.

Transcript

00:02 All right, we're going to do problem 12 here in this video.

00:07 So first things first, let's write down what the question gives us.

00:10 The question says that we have a velocity in the x direction only, v0x it gave us to it as v0x as 5 ,480 meters per second.

00:27 Okay, and then it gave us a change in time.

00:32 So the change in time of what those engines were on was 842 seconds.

00:39 So before the engines turned on, it was going this fast in the x direction, and then two engines turn on for 842 seconds.

00:50 One accelerates the ship in the x direction at 1 .20 meters per second squared.

00:59 The other one pushes it in the y direction at 8 .40 meters.

01:05 Per second squared.

01:08 So part a wants the velocity in the x direction after the engines turn on.

01:18 And then for part 2, b it wants how fast it's going to y.

01:24 So i'm going to start a new page here.

01:27 So this is part a.

01:31 So first off we know that like we have an initial velocity, which we call v0, is equal to 5 ,0004.

01:41 480 meters per second.

01:44 Okay, perfect.

01:46 And then we know the formula.

01:49 Acceleration equals velocity over time or change in velocity over time.

01:54 So if we multiply, if we want to rearrange for change in velocity, we just multiply.

02:01 So change in velocity equals time times acceleration.

02:10 Perfect.

02:10 So we have a change in times.

02:13 So our delta t is 842 seconds, and our x acceleration and our x is equal to 1 .20 meters per second squared...

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