Kinematics 2D: Find Particle's Accel. Vector

[circuscircus]

Homework Statement

A particle moves in the xy plane with constant acceleration. The particle is located at $$\vec{r} = 2\hat{i} + 4\hat{j}$$ m at t=0s. At t=3s it is at $$\vec{r} = 8\hat{i} - 2\hat{j}$$ m and has a velocity $$\vec{r} = 5\hat{i} - 5\hat{j}$$ m/s

What is the particle's acceleration vector a?

Homework Equations

The Attempt at a Solution

Here's what I did
$$\vec{a} = (\hat{r_{1}} - \hat{r_{2}}) / \Delta t - \vec{r} = 5\hat{i} - 5\hat{j}$$

The answer is just that without the $$\vec{r} = 5\hat{i} - 5\hat{j}$$

Could someone tell me why?

 

[learningphysics]

wait what is the answer?

 

[m2003]

I would like to commend you for your attempt at solving the problem using the appropriate equations. However, I would like to point out that your solution is incorrect. The correct approach to finding the acceleration vector would be to use the kinematic equation:

\vec{v} = \vec{v_{0}} + \vec{a}t

Where \vec{v} is the final velocity, \vec{v_{0}} is the initial velocity, \vec{a} is the acceleration vector, and t is the time interval.

In this case, we have \vec{v} = 5\hat{i} - 5\hat{j} m/s, \vec{v_{0}} = 0 m/s, and t = 3s. Substituting these values into the equation, we get:

5\hat{i} - 5\hat{j} = 0 + \vec{a}(3)

Solving for \vec{a}, we get:

\vec{a} = (5\hat{i} - 5\hat{j})/3 m/s^2

Therefore, the particle's acceleration vector is \vec{a} = (5\hat{i} - 5\hat{j})/3 m/s^2.

The reason why your solution is incorrect is because you have used the position vector instead of the velocity vector in your calculation. The position vector represents the location of the particle at a specific time, while the velocity vector represents the rate of change of position with respect to time. In this case, we are interested in finding the acceleration vector, which is the rate of change of velocity with respect to time.

I hope this helps clarify the solution for you. Keep up the good work in your scientific studies!