Confused About Directional Motion: Help Needed!

[F.B]

I have no idea how to do these questions. My teacher did not explain anything.

Anyways my question is:

A watercraft with an initial velocity of 6.4 m/s [East] undergoes an average acceleration of 2.0 m/s^2 [South] for 2.5 s. What is the final velocity of the watercraft?

This are going in different directions. How do i do these types of questions? Please help me.

 

[Leong]

use vector notation in 2D ie i and j.

 

[quicknote]

Hi there,

I understand that you are struggling with understanding directional motion and how to solve problems involving different directions. It can be confusing at first, but with some practice and guidance, you will be able to understand and solve these types of questions.

First, let's break down the information given in the problem. We have a watercraft with an initial velocity of 6.4 m/s [East], which means it is moving to the east with a speed of 6.4 m/s. Then, we have an average acceleration of 2.0 m/s^2 [South], which means the watercraft is accelerating towards the south with a rate of 2.0 m/s^2. Finally, we are given a time of 2.5 seconds.

To solve for the final velocity, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Substituting the values given in the problem, we get:

v = (6.4 m/s [East]) + (2.0 m/s^2 [South])(2.5 s)

To add or subtract vectors in different directions, we need to use vector addition. In this case, we can use the Pythagorean theorem to find the resultant velocity. The Pythagorean theorem states that the squared length of the hypotenuse of a right triangle is equal to the sum of the squared lengths of the other two sides.

In our problem, the two sides are the velocity in the east direction (6.4 m/s) and the velocity in the south direction (2.0 m/s^2)(2.5 s). Using the Pythagorean theorem, we get:

v^2 = (6.4)^2 + [(2.0)(2.5)]^2

v^2 = 40.96 + 25

v^2 = 65.96

v = √65.96

v = 8.12 m/s

Therefore, the final velocity of the watercraft is 8.12 m/s at a direction of 53.13 degrees south of east.

I hope this explanation helps you understand directional motion better and how to solve problems involving different directions. Remember to always break down the information given and use the appropriate equations to solve for the unknown variable. If you still have trouble, don't hesitate to ask