Finding Acceleration in a Moving Particle

[habibclan]

Homework Statement

A particle moves in the xy-plane with constant acceleration. The particle is located at r= (2i + 4j)m at t=0 s. At t=3 s it is at r=(8i - 2j) m and has a velocity v= ( 5i - 5j) m/s.
a) What is the particle's acceleration vector a?
b) What are its position, velocity and speed at t= 5s?

Homework Equations

xf= xo + vt+ 0.5 a t^2
yf= yo + vt+ 0.5 a t^2

The Attempt at a Solution

I used the above two equations to solve for the acceleration in both x and y components separately and I got a= (1.33i - 1.33j), but the answer at the back of the textbook is a= (2i - 2j). I don't understand what I'm doing wrong. Can someone guide me please? Thanks in advance!

 

[dirk_mec1]

Show us the steps you took to get the 'wrong' answer.

 

[alphysicist]

Hi habibclan,

In your equations, the v is the initial velocity, and it appears that you plugged in 0 for that. However, this particle does not start at rest, and the initial velocity is unknown. (So your x equation, for example, has two unknowns--the initial velocity in the x direction and the acceleration in the x direction.)

However, you do also know the final velocity in the x direction. Do you see what to do with that?

 

[habibclan]

For the equation, i plug in the initial and final x-component position, the initial velocity as 0, and t=3 s and solve for a.

xf= xo + vt+ 0.5 a t^2
8 = 2 + 0.5*a * (3^2)
a= 1.33

Therefore, the acceleration of the x-component is 1.33 m/s^2

For this one, its the same as above, except i plug in the intial and final y-comp. positions.

yf= yo + vt+ 0.5 a t^2
-2=4 + 0.5*a* (3^2)
a= -1.33

Therefore, acceleration of the y-comp. is -1.33 m/s^2.

What am I doing wrong please?

 

[dirk_mec1]

The initial velocity is unknown you'll have to solve 2 equations with two unknowns.

 

[habibclan]

Hi habibclan,

In your equations, the v is the initial velocity, and it appears that you plugged in 0 for that. However, this particle does not start at rest, and the initial velocity is unknown. (So your x equation, for example, has two unknowns.)

However, you do also know the final velocity in the x direction. Do you see what to do with that?

I got it! Thank you so much! I used the equation vf= vi + at to solve for initial velocity and then plugged it in. Stupid of me to assume that the initial velocity was 0. Thanks a lot!