(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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, F_{1}and F_{2}act upon the object

F_{1}= [1 N + (1/2 N/s)t ][tex]\hat{i}[/tex]

F_{2}= -(1/3 N/s^{2})t^{2}[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?

2. Relevant equations

F=ma

kinematics

3. 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/s^{2})t^{2}[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]_{j}t

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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# Homework Help: Confirm correctness of simple newton's second law-type problem?

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