Physics Olympiad Vector Acceleration

[bazookajason]

Homework Statement

A particle of mass 2.00 kg moves under a force given by
F~ = −(8.00 N/m)(xˆi + yˆj)
whereˆi and ˆj are unit vectors in the x and y directions. The particle is placed at the origin with an initial velocity~v = (3.00 m/s)ˆi + (4.00 m/s)ˆj.

a. After how much time will the particle first return to the origin?
b. What is the maximum distance between the particle and the origin?

Homework Equations

f=ma
x=vt+1/2at^2

The Attempt at a Solution

Using f=ma, i find the acceleration to be a=-4x i + -4y j
Using x=vt+1/2at^2, if i set x as 0
So i get 0=3t+1/2(-4x)t^2 and 0=4t+1/2(-4y)t^2
I solve those 2 equtaions to get 3y=4x
but i don't know where to continue

 

[SammyS]

Homework Statement

A particle of mass 2.00 kg moves under a force given by
F~ = −(8.00 N/m)(xˆi + yˆj)
whereˆi and ˆj are unit vectors in the x and y directions. The particle is placed at the origin with an initial velocity~v = (3.00 m/s)ˆi + (4.00 m/s)ˆj.

a. After how much time will the particle first return to the origin?
b. What is the maximum distance between the particle and the origin?

Homework Equations

f=ma
x=vt+1/2at^2

The Attempt at a Solution

Using f=ma, i find the acceleration to be a=-4x i + -4y j
Using x=vt+1/2at^2, if i set x as 0
So i get 0=3t+1/2(-4x)t^2 and 0=4t+1/2(-4y)t^2
I solve those 2 equations to get 3y=4x
but i don't know where to continue

The equation

x=vt+(1/2)at²​

has 2 big problems.

1. It's only true for uniform (constant) acceleration. The acceleration here is not constant.

2. Even if the acceleration were constant you should only have included x components.​

 

[bazookajason]

hm do i use calculus?
a= derivative of x

 

[bazookajason]

anyone?

 

[asteriatic]

from here.

First, you have correctly calculated the acceleration of the particle to be a = -4x ˆi + -4y ˆj. To find the time it takes for the particle to return to the origin, you 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. Since the final velocity is zero when the particle returns to the origin, you can set v = 0 and solve for t. This will give you the time it takes for the particle to return to the origin.

To find the maximum distance between the particle and the origin, you can use the equation s = ut + 1/2at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. You can use the same value of t that you calculated in the previous part to find the maximum distance. This will give you the maximum displacement of the particle from the origin.