How to Use Integral to Solve Problems

[Stephanus]

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?

$V=dx/dt$ then $dt=dx/V$ then integrate it $V=HD$ İnside the integral will be $1/H_0xdx$

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/s²
For a given time, how can we find the distance it takes?
I know the answer is D = 1/2 at². We were taught that at high school.
Instinctively it is the integral of D = VT
I'll try to make a sentence like RyanH42.
---------------------------------------------------------------------------------------------------------------------------------------------
$V = dx/dt$ then $dx = V dt$ then integrate it $V = a.t$ inside the integral will be $\int a.t \, dt$
--------------------------------------------------------------------------------------------------------------------------------------------
Is this sentence correct? I try to understand how to use the integral to solve a problem not to integrate an equation.

 

[jedishrfu]

The "at" expression in the integral ∫a.tdt comes from an inner integral ∫adt = at is that what you mean?

 

[RyanH42]

$V=dx/dt$
$dx=Vdt$
$x=∫Vdt$
$a=dV/dt$
$dV=adt$
$V=∫adt$ 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 $10t^2/2$

 

[Mark44]

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.

This doesn't make sense. Acceleration is the time rate of change of velocity, so the units typically are m/s² or similar (distance/time²).

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/s²
For a given time, how can we find the distance it takes?
I know the answer is D = 1/2 at². We were taught that at high school.
Instinctively it is the integral of D = VT
I'll try to make a sentence like RyanH42.
---------------------------------------------------------------------------------------------------------------------------------------------
$V = dx/dt$ then $dx = V dt$ then integrate it $V = a.t$ inside the integral will be $\int a.t \, dt$
--------------------------------------------------------------------------------------------------------------------------------------------
Is this sentence correct? I try to understand how to use the integral to solve a problem not to integrate an equation.

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 v₀ and s₀ are the initial velocity and initial displacement.

 

[Stephanus]

The "at" expression in the integral ∫a.tdt comes from an inner integral ∫adt = at is that what you mean?

I mean Distance = $\int a.t \, dt$
But if you said

∫adt = at

Is it $V = \int a \, dt$?

 

[Stephanus]

$V=dx/dt$
$dx=Vdt$
$x=∫Vdt$
$a=dV/dt$
$dV=adt$
$V=∫adt$ 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 $10t^2/2$

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 $\int a \, dt = a.t \text { , so } D = \int a.t \, dt$

 

[Stephanus]

Hi Mark44

This doesn't make sense. Acceleration is the time rate of change of velocity, so the units typically are m/s² or similar (distance/time²).

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.

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 v₀ and s₀ are the initial velocity and initial displacement.

Yes it makes sense.
It is for the case where the object started not in T₀ 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.