A fish swimming in a horizontal plane has velocity 𝐯i=(4.00 𝐢+1.00 𝐣) m / s at a point in the oc

A fish swimming in a horizontal plane has velocity 𝐯i=(4.00...

A fish swimming in a horizontal plane has velocity $\mathbf{v}_{i}=(4.00 \mathbf{i}+1.00 \mathbf{j}) \mathrm{m} / \mathrm{s}$ at a point in the ocean where the position relative to a certain rock is $\mathbf{r}_{i}=(10.0 \hat{\mathbf{i}}-4.00 \hat{\mathbf{j}}) \mathrm{m}$ After the fish swims with constant acceleration for $20.0 \mathrm{s}$ its velocity is $\mathbf{v}=(20.0 \hat{\mathbf{i}}-5.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} .$ (a) What are the components of the acceleration? (b) What is the direction of the acceleration with respect to unit vector $\hat{\mathbf{i}} ?$ (c) If the fish maintains constant acceleration, where is it at $t=25.0 \mathrm{s},$ and in what direction is it moving?

A fish swimming in a horizontal plane has velocity $\mathbf{v}_{i}=(4.00 \mathbf{i}+1.00 \mathbf{j}) \mathrm{m} / \mathrm{s}$ at a point in the ocean where the
position relative to a certain rock is $\mathbf{r}_{i}=(10.0 \hat{\mathbf{i}}-4.00 \hat{\mathbf{j}}) \mathrm{m}$
After the fish swims with constant acceleration for $20.0 \mathrm{s}$ its velocity is $\mathbf{v}=(20.0 \hat{\mathbf{i}}-5.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} .$ (a) What are the
components of the acceleration? (b) What is the direction of the acceleration with respect to unit vector $\hat{\mathbf{i}} ?$ (c) If the fish maintains constant acceleration, where is it at $t=25.0 \mathrm{s},$ and in what direction is it moving?

Key Concepts

Determining Vector Direction

Determining the direction of a vector involves computing the angle it makes with a reference axis, often using trigonometric functions. This method is crucial for understanding the relative orientation of velocity and acceleration vectors, which in turn helps in predicting the path of an object in two-dimensional space.

Kinematic Equations

Kinematic equations under constant acceleration relate displacement, initial velocity, final velocity, acceleration, and time. They provide a mathematical framework to calculate unknown quantities in motion problems, and are especially useful in predicting an object's position and velocity at any given time.

Vector Kinematics

This concept involves representing physical quantities such as displacement, velocity, and acceleration as vectors that have both magnitude and direction. It forms the basis for analyzing motion in two dimensions, allowing the separation of complex movement into simpler, independent components.

Constant Acceleration

Constant acceleration implies that the change in velocity remains uniform over time. This simplifies the analysis using kinematic equations since the acceleration vector remains unchanged, which is fundamental in predicting future positions and velocities of moving objects.

Decomposition of Vectors

Decomposing vectors into their components (typically along the horizontal and vertical axes) allows for the independent analysis of each direction. This technique is essential when working with motion in a plane because it enables solving problems using simple scalar equations for each dimension.

Transcript

00:01 All right, in this kinematics problem, the initial position of a fish in two dimensions is given.

00:06 We also have the initial velocity of a fish and the final velocity of the fish.

00:10 We know that in between the initial and the final velocity, a time passes of t equals 20 .0 seconds.

00:16 And we know that a equals a constant value over that time.

00:22 So given that, we need to find the components of a and the i and j direction, and then find the direction of a.

00:33 All right? so first, we need to remember that if a is constant, then either the x or y direction of a can be found by calculating the change of the velocity in just that direction divided by the change in time.

00:50 So in this case, the final x velocity minus the initial x velocity divided by the change in time will give us the x acceleration.

01:00 All right.

01:01 So we can take that formula and we can apply that to the i -hat direction and the j -hat direction separately.

01:08 So the x and the y -dre direction separately.

01:10 In this case, the i -hat direction with bf minus v i -mi is 20 meters per second minus four meters per second divided by delta t 20 seconds in the i -hat direction.

01:21 And this is plus for the j -hat direction minus five meters per second, the final, minus one meter per second, the initial divide.

01:30 By 20 seconds and this is j hat direction and our units are meters per second square all right so if we calculate out this whole thing it becomes 0 .8 0 in the i hat direction minus 0 .30 in the j hat direction measured meters per second squared all right so that those are the components of the acceleration in the j hat direction measured meters per second squared all right so that those are the components of the acceleration in the i -hat 0 .8 -0 and 0 .30 for j -hat.

02:06 We can keep three significant figures for both of that because all of our values up here have at least three significant figures.

02:13 All right.

02:14 To find the direction of a, we need to calculate theta using a tan inverse function.

02:25 So for a tan inverse function, if theta is measured from the x -axis, then we're going to put the value of our y component on top, so y component of acceleration on top, and x component on bottom, and then we can calculate.

02:42 In our calculators, this will give us about 339 degrees, or if it'll give you a negative value, it'll give you about negative 21 degrees.

02:52 All right.

02:52 So, i mean, that makes sense because in our x value, we have over here a positive i and a negative j hat value so our if this is i hat our a will be going downwards to the right like that all right so there you have for the first part the components of a and the direction of a right next your question asks what is the new position are at t equals 25 .0 seconds if a is constant and if a stays this constant value the entire time.

03:31 So find the new position r at t equals 25 .0 seconds and find the direction of motion.

03:38 So the direction of the v, the v vector, all right, our velocity vector.

03:48 So for the position, we need to find equations for our x position and our y position as functions of time separately.

03:56 For exposition, we have x of t is initial exposition, x not, plus initial x velocity times time, plus one half x acceleration times time squared.

04:13 Don't forget the squared.

04:14 And this will give you for our problem, 10 meters plus 4 meters per second times 25 seconds plus 1⁄2 .8 meters per second...

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