Two D kinematics

1. Oct 21, 2007

Grapz

1. The problem statement, all variables and given/known data
A particle moves in the xy-plane with constant acceleration. The particle is located at r= 2i + 4j at t = 0 s. At t = 3s it is at r = 8i - 2j and has velocity v = 5i - 5j

What is the particle's acceleration vector

2. Relevant equations

X_f = x_i + v_ix(delta t) + 1/2 a_x (delta t )^2 (1)
Y_f = y_i + v_iy(delta t) + 1/2 a_y (delta t )^2 (2)

3. The attempt at a solution

So i try to solve for a_x and a_y individually.

since t = 3, xi = 2, xf = 8

i use formula #1. Now i come across a problem, the question does not give me the initial velocity, and i tried to assume that it was 0, but when i did that it doesnt' match the answer in the book. So now i am really confused as to how to solve this problem. The same occured when i tried to solve for the acceleration in the y direction.

How do i solve for v initial in both x and y direction?

the answer is a = 2i - 2j

Last edited: Oct 21, 2007
2. Oct 21, 2007

learningphysics

Use the fact that velocity at t = 3 is v = 5i - 5j

X_f = x_i + v_ix(delta t) + 1/2 a_x (delta t )^2 (1)

8 = 2 + v_ix(3) + 1/2a_x(3)^2 equation 1

You also know that

V_f = v_ix + a_x*t

5 = v_ix + a_x(3) equation 2

use equations 1 and 2 to solve for a_x.

You can also more directly for a_x solve using

X_f = x_i + v_f(delta t) - 1/2 a_x (delta t)^2

this formula is less commonly presented in texts...

3. Oct 21, 2007

Grapz

Ah i forgot about that equation.

Thank you :P