(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity

(a) In what direction would the ship in Exercise 3.57 have...

Key Concepts

Relative Velocity

Relative velocity is the concept of analyzing the motion of an object with respect to another moving reference frame. In problems involving objects moving in mediums such as water or air, the observed velocity is the vector sum of the object's velocity relative to the medium and the medium's velocity relative to the observer.

Vector Addition

Vector addition involves combining two or more vectors to determine a resultant vector. In relative motion scenarios, this method is used to add the velocities of the ship and the current to obtain the net velocity relative to the Earth.

Component Analysis

Component analysis is the process of breaking vectors into their horizontal and vertical components. By resolving vectors into perpendicular components, especially using trigonometric functions, one can analyze each direction independently, which simplifies the calculation of the resultant motion.

Trigonometric Resolution

Trigonometric resolution involves using trigonometric ratios to determine the magnitude of vector components when the vector's magnitude and the angle of motion are known. This technique is essential for finding the direction in which the ship must travel to counteract the current and achieve a desired net velocity.

Transcript

00:02 In a previous example, we looked at a boat that was traveling straight due north relative to the water it was in.

00:12 But relative to the earth, it was traveling at an angle like this because there was a current in the water that pointed up like this at an angle of 40 degrees north of east.

00:27 And we determined what the velocity of our boat was relative to the earth by summing these two vectors together component -wise and then bringing them back at the end to get the final magnitude.

00:44 So now what we want to know, and sorry, not these two vectors, but this vector and this vector to get this vector.

00:55 And now what we want to know is what would this vector on the left have to be in order for our boat to just go due north.

01:07 So if the boat's going to go due north, we know that this vector, the velocity of our boat, will probably have to point to the left a bit like this in order to cancel out the x component from the current here, which is precisely what i've drawn up here.

01:25 So specifically what we want to figure out is what angle would our boat have to move at relative to the water in order for us to just move straight due north relative to earth.

01:40 So we already calculated what the components of velocity of the water are relative to the earth.

01:47 So those are right here.

01:50 And we know that the velocity of our boat is going to be seven meters per second regardless of which direction it's pointing.

01:56 And i just said a moment ago that the condition in our problem that will make us go straight due north as if the velocity of our boat in the x direction relative to the water is equal and opposite to the velocity of the water relative to the earth in the x direction.

02:25 So we want the component of velocity in the x direction for our boat relative to the water to be minus 1 .15 meters per second.

02:44 Okay.

02:46 So that will make it so that the x components cancel out.

02:51 So if that makes it so the x components cancel out, we can determine now what our final magnitude of velocity should be.

02:59 And the angle.

03:01 For the first part, we're just going to look at the angle.

03:05 And we can do this by just using our trigonometric relationships between the magnitude of our vector and its components.

03:13 And that's just going to be that the x component of our vector will be equal to its magnitude times the cosine of the angle it makes with the horizontal.

03:50 Okay.

03:54 So we can go ahead and solve this for theta.

03:57 Just by dividing both sides and taking the inverse cosine.

04:04 And we'll have the theta is equal to the cosinverse of minus 1 .15 meters per second divided by 7 meters per second.

04:17 So this should give us this angle here.

04:22 There's this another solution for the arc cosine that would be when this vector is pointing in this direction.

04:28 But we definitely aren't concerned with that since.....

Please give Ace some feedback