/sci/ - Science & Math

[Anonymous No.1030166]

I'm curious, where do we get the formula for distance traveled of a constantly accelerating object? I've tried figuring it out, but no luck so far. It's really starting to bother me.

Pic related, it's the formula. No, this has nothing to do with homework, just my own curiosity.

[Anonymous No.1030171]

You need the other equations, faget.

[Anonymous No.1030191]

>>1030171

What equations?

[Anonymous No.1030229]

>>1030166
>>1030166

From integration

<div class="math">a=\frac{\mathrm{d} v}{\mathrm{d} t} ;v=\int adt=at+C; v=\frac{\mathrm{d} s}{\mathrm{d} t};s=\int vdt=\int (at+C)dt=D+Ct+\frac{1}{2}at^2</div>

D is a constant determined by initial position, C is initial speed

[Anonymous No.1030278]

>>1030229

Thank you very much, kind sir. Though it is currently beyond my understanding in mathematics, it does give me some insight into how it works.

I'm trying to teach myself some math and physics out of interest and so I can hopefully skip some of the more basic classes when I go back to college next semester.

[Anonymous No.1030293]

>>1030278
you won't be able to.

[Anonymous No.1030330]

>>1030293

Thanks.

[Anonymous No.1030396]

Consier with a starting velocity u=o. The distance now = (1/2)at^2. Think of the average velocity of the particle = (0 + at)/2 =at/2. Distance is just velocity times time ie (at/2)*t = 1/2at^2. The ut term accounts for the additional distance travelled due to a non zero starting velocity. Another way to look at it is to plot a graph of velocity against time for a constantly accelerating object. The area of the graph is equal to the distance travelled. It will be a trapezoid if u is not 0, and putting the variables into the trapezoid area formula will yield the same expression.

[Anonymous No.1030423]

Look at the defenition of acceleration, a = (delta(velocity))/(delta(time)), that is, how much does the velocity change over a certain time.

Let's look at the same thing, tho in infinitley small steps: a = dv/dt. From here, toss up that dt and integrate both sides. a*dt = dv (dv: from v`-initial to v-final, dt: from 0 to t) => a*t = v-v` <=> v = v` + at. Now we know that v = dx/dt, so toss up that dt and integrate once more and you'll land in something like distance with respect to time.

[Anonymous No.1030430]

learn to calculus.

[rec & play !!ZTOm5LLVLJH No.1030465]

My graph shows speed vs time. Area of my graph (grey) will be distance. Work out the area using basic geometry, and there's your simple intuitive understanding.

[Anonymous No.1030517]

>>1030465

I like this, a lot. Thanks.

[Anonymous No.1030539]

You need to understand integration for this.