Dear PF Forum,
How to use integral?
I created a thread months ago to find out what is the integral of Hubble equation.
Now, I create a thread to understand how to use the integral.

Problem 1:
Something is constantly accelerating 10m/s per meter. Let's say the acceleration is H.
so H = 10m . s^{-1}. h^{-1}
Velocity = H x Distance; V = HD
For a given distance, how can we find the time it takes to cover that distance?

Someone has given me the answer. Okay..., I get it.

Problem 2:
Something is constantly accelerating 10m/s per second. Let's say the acceleration is a
so a = 10 m.s^{-1}.s^{-1}
or a = 10 m/s^{2}
For a given time, how can we find the distance it takes?
I know the answer is D = 1/2 at^{2}. We were taught that at high school.
Instinctively it is the integral of D = VT
I'll try to make a sentence like RyanH42.
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##V = dx/dt## then ##dx = V dt## then integrate it ##V = a.t## inside the integral will be ##\int a.t \, dt##
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Is this sentence correct? I try to understand how to use the integral to solve a problem not to integrate an equation.

##V=dx/dt##
##dx=Vdt##
## x=∫Vdt##
##a=dV/dt##
##dV=adt##
##V=∫adt## so

##x=∫Vdt## here V is equal ##V=∫adt## = ##x=∫(∫adt)dt## a is 10 so, ##V=∫adt ##, ##V=∫10dt ## which its 10t. now ##x=∫(∫adt)dt##=##x=∫10tdt## which its ##10t^2/2##

This doesn't make sense. Acceleration is the time rate of change of velocity, so the units typically are m/s^{2} or similar (distance/time^{2}).

To solve this problem you have to solve a differential equation, which in this case you can do by integrating twice.
The differential equation is ##\frac{d^2s}{dt^2} = a##.
Integrate both sides to get ds/dt, the velocity:
Velocity: ##v = \frac{ds}{dt} = \int \frac{d}{dt}\frac{ds}{dt} dt= \int a dt = at + v_0##
Integrate once more to get the displacement:
Displacement ##s = \int \frac{ds}{dt} dt= \int v dt = \int (at + v_0)dt = \frac 1 2 at^2 + v_0t + s_0##
The constants v_{0} and s_{0} are the initial velocity and initial displacement.

Ahh, this is what @jedishrfu means about inner integral?
So the distance is not straight forward ##D = \int a.t \, dt##, but ##D = \int (\int a \, dt) \, dt## while ##\int a \, dt = a.t \text { , so } D = \int a.t \, dt##

YES, That's right Mark44, it's NOT acceleration. You might know that the galaxies are traveling from us. The farther, the faster. About 100km/s per 3 million light year. (200 km/s per 6 million ly) You are right! It's not acceleration. But for a non English speaker. I don't know the term for the galaxy movement. Glad you pointed that out.

Yes it makes sense.
It is for the case where the object started not in T_{0} and at some distance at some velocity. Have to think this paragraph for a while to understand that. It's more complicated than I tought.