/sci/ - Science & Math

[Anonymous No.8340113]

I know this be taken as a homework request, but it isn't really

I'm 38 now, I haven't done any math since I left highschool, and I don't remember anything, I've been working in IT most of my adult life, and never used any math skills. Now I've signed up for an electronics course, and it is asking me a question like this for basic math review,

>For the object with zero initial velocity, suppose its acceleration is illustrated in the figure. Its velocity v(3s) = _ m/s

Now, can someone point me to what I need to learn in order to solve this? I don't want you to solve it, but if you could explain what kind of question this is and perhaps the relevant syllabus on khan academy or similar I'd be quite grateful.

[Anonymous No.8340119]

Integral calculus

[Anonymous No.8340121]

>>8340113
Look up the 5 kinematic equations and look for v_f = v_i + a*t

[Anonymous No.8340125]

>>8340113
I think you need calculus to solve this, or it makes it much easier, since acceleration varies with time. You can't use your regular kinematic equations since those assume constant acceleration.

What you need to know is that acceleration is the derivative of velocity, so all you need to do is find the equation of the curve in your pic, then integrate from 0 to 3. A more simple way is to know that an integral gives you the area under the curve, so all you need to do is find the area under the red line. Hint: It's a trapezoid.

[Anonymous No.8340143]

>>8340113
You could either find the area under the curve by splitting it into a rectangle and a triangle... Or you can use v = v0 + at with v0=0 and a =2 and t =3

[Anonymous No.8340150]

>>8340113
Calculus problem you can solve with geometry.

Calculate the area of the trapezoid formed by the red line, and the "a" and "t" axes. You could do this by dividing up the shape into two: a square and a triangle. Add up the two areas and you have your velocity.

[Anonymous No.8340151]

>>8340121
>>8340143
You can't use kinematic equations for non-constant acceleration.

[Anonymous No.8340158]

>>8340151
my b forgot havent done physics since hs xD

[Anonymous No.8340257]

Thanks goys!

[Anonymous No.8340317]

>>8340113
[math]\int_{0}^{3}{\frac{1}{3}x+1 dx}[/math]
as
[math]v = \frac{dv}{dt}\\and\\a = \frac{d^{2}v}{dt^{2}}[/math]

[Anonymous No.8340325]

[math] \displaystyle
position \; \frac{ \overset{/time}{ \rightarrow}}{ \underset{ \texttt{x} \, time}{ \leftarrow}} \; speed \; \frac{ \overset{/time}{ \rightarrow}}{ \underset{\texttt{x} \, time}{ \leftarrow}} \; acceleration
[/math]