Confirm correctness of simple Newton's second law-type problem?

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zero_infinity
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Homework Statement


at t=0, a 4kg mass is moving at 2 m/s[tex]\hat{i}[/tex] + 3m/s[tex]\hat{j}[/tex]
at t=0, F1 and F2 act upon the object
F1= [1 N + (1/2 N/s)t ][tex]\hat{i}[/tex]
F2= -(1/3 N/s2)t2 [tex]\hat{j}[/tex]

a) what is the object's velocity when t=3?
b) at what time does the object stop moving in [tex]\hat{j}[/tex] direction?
c) what is the object's displacement when t=6?

Homework Equations


F=ma
kinematics

The Attempt at a Solution


just look at the steps and see if I'm doing it right please

[tex]\vec{V}[/tex]0 = 2 m/s[tex]\hat{i}[/tex] + 3m/s[tex]\hat{j}[/tex]
[tex]\vec{F}[/tex](t) = [1 N + (1/2 N/s)t ][tex]\hat{i}[/tex] - (1/3 N/s2)t2 [tex]\hat{j}[/tex]


(a)
[tex]\vec{F}[/tex](t) = m[tex]\vec{a}[/tex]
[tex]\vec{F}[/tex](3)/4 kg = [tex]\vec{a}[/tex]

Use [tex]\vec{V}[/tex]f = [tex]\vec{V}[/tex]0 + [tex]\vec{a}[/tex]t

[tex]\vec{V}[/tex]f = 21/8 [tex]\hat{i}[/tex] - 3/4 [tex]\hat{j}[/tex]





(b)
just look at the [tex]\hat{j}[/tex] component of the vector

[tex]\vec{F}[/tex]j(t)/ 4kg = [tex]\vec{a}[/tex]j
Use [tex]\vec{V}[/tex]f = [tex]\vec{V}[/tex]0 + [tex]\vec{a}[/tex]t

0 = 3m/s[tex]\hat{j}[/tex] - [tex]\vec{a}[/tex]jt

t = 3.30 s





(c)
[tex]\vec{F}[/tex](t)/ 4kg = [tex]\vec{a}[/tex](t)

actually, now when I'm rewriting it, I'm not sure how to integrate [tex]\vec{a}[/tex]

do i integrate the acceleration from 0 to 6? i don't think so...

i integrated acceleration (not accounting for C) and then integrated velocity from 0 to 6 (this is obviously wrong now that I'm looking over it)
 
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Find the velocity and acceleration along i and j directions separately.
Using proper kinematic equations find the remaining answers by using vector addition.
 
rl.bhat said:
Find the velocity and acceleration along i and j directions separately.
Using proper kinematic equations find the remaining answers by using vector addition.

Hi

which part are you talking about?