average with respect to time vs with respect to distance

mean average with respect to time vs with respect to distance

Notation

I am writing 2*(1second)^2 with a shorthand of 2 second ^ 2 that is why I am writing meter not meters and second not seconds.  I am doing this because it became very confusing writing out the units differently for plural and singular form such as 2 seconds but 1 second when the units were put to powers other than 1.

x = displacement

t = time

m = mass

f = force

a = acceleration

v = velocity

P = momentum

KE = Kinetic Energy

Int[y, g(y)] = the anti derivative of g(y) with respect to y

dg(y)/dy = the derivative of g(y) with respect to y

Dyg(y) = the derivative of g(y) with respect to y

Example 1

given x(t=0 seconds) = 0 meter

given v(t=0 seconds) = 0 meter second ^ -1

given a(t) = 1 meter second ^ -2

v(t) = t*(1 meter second ^ -2)

x(t) = 0.5 meter second ^ -2 * t^2

t^2 = x / (0.5 meters * seconds ^ -2 ) 

t^2 =  2*x / ( 1 meter per second ^ 2 )

t^2'= (2 meter^-1 second ^2) * x

t(x) = (2^0.5)*(1 meter ^ -0.5 second ^ 1) *(x^0.5)

v(x) = t(x) * (1 meter second ^ -2)

v(x) = (2^0.5)*(1 meter ^ -0.5 second ^ 1) *(x^0.5) * (1 meter second ^ -2)

v(x) = (2^0.5)*(1 meter ^ 0.5 second ^ -1) *(x^0.5)

antiderivative of v(x) with respect to x shall be called Int[x, v(x)]

Dx(x^N) = N*x^(N-1)

Int(x, x^N) = (1/[N+1])*x^(N+1) + K0

Dx(1/[N+1])*x^(N+1) = (N+1)/(N+1)*x^(N+1-1)

Int[x, v(x)] = (1/1.5)*(x^1.5)*(2^0.5)*(1 meter ^ 0.5 second ^ -1) + K1

Int[x, v(x)] = (2/3)*(x^1.5)*(2^0.5)*(1 meter ^ 0.5 second ^ -1) + K1

Int[x, v(x)] = (x^1.5)*(2^1.5)*(1 meter ^ 0.5 second ^ -1) / 3 + K1

x(t) = 0.5 meter second ^ -2 * t^2

x (1 second) = 0.5 meter second ^ -2 * (1 second) ^ 2 = 0.5 meter

Int[x, v(x= 0.5 meter)] = (0.5 meter^1.5)*(2^1.5)*(1 meter ^ 0.5 second ^ -1) / 3 + K1

0.5 ^ 1.5 * 2 ^ 1.5 = 1

Int[x, v(x=0.5 meter)] = ( 1 meter ^2 second ^ -1 ) / 3 + K1

Int[x, v(x=0 meter)] = ( 0 meter ^2 second ^ -1 ) / 3 + K1

Average velocity with respect to displacement from a displacement of 0 meters to a position of 0.5 meter =

= ( Int[x, v(x= 0.5 meter)] - Int[x, v(x=0 meter)] ) / ( 0.5 meter - 0 meter) =

= ( 1 meter ^ 2 / 3 ) / 0.5 meter =

= 2 meter / 3

antiderivative of v(t) with respect to t shall be called Int[t, v(t)]

v(t) = t*(1 meter second ^ -2)

Int[t, v(t)] = 0.5 meter second ^ -2 * t^2 + K2

Int[t, v(t=1 second) = 0.5 meter second ^ -2 * (1 second )^2 + K2

Int[t, v(t=1 second) = 0.5 meter + K2

Int[t, v(t=0 second) = 0 meter + K2

Average velocity with respect to time from a time of 0 second to a time of 0.5 second =

= ( Int[x, v(t= 1 second)] - Int[x, v(t=0 second)] ) / ( 1 second - 0 second) =

= (0.5 meter-0meter) / (1 second-0second) = 0.5 meter second ^ -1

given mass = 1 kilogram

KE(x) = 0.5 * 1 kilogram * [v(x)]^2

v(x) = (2^0.5)*(1 meter ^ 0.5 second ^ -1) *(x^0.5)

KE(x) = 1 kilogram meter second ^ -2 * x

F(x) = dKE/dx = 1 Kilogram meter second ^ -2

P(t) = 1 kilogram * v(t)

v(t) = t*(1 meter second ^ -2)

P(t) = t*(1 kilogram meter second ^ -2)

F(t) = dP(t)/dt = 1 kilogram meter second ^ -2

The average velocity with respect to time is not the same as the average velocity with respect to displacement for an object that starts stationary and experiences constant acceleration when averaged over a distance interval that is the same distance as it would have traveled in the time interval in which it's average velocity with respect to time is calculated.  However the average force it would experience over such an interval would be the same if either averaged by time or by displacement because the force is the same constant amount in either case.  Such would hold true for a falling object experiencing a constant force having displacement, time and velocity measured as zero at the initial starting condition of the fall.

Example 2

given a(t) = t^2 * 1 meter second ^ -4

given v(t=0 second) = 0 meter second ^ -1

given m = 1 kilogram

given x(t=0 second) = 0 meter

v(t) = 4*t^3 * 1 meter second ^ -4

x(t) = 12*t^4* 1 meter second ^ -4

[t(x)] ^ 4 = x / 12 meter second ^ -4

[t(x)] ^ 4 = x*(1/12)*1 meter ^ -1 second ^ 4

t(x) = x^(0.25) * (1/12)^0.25 * 1 meter^-0.25 second ^1

v(x) = 4 * 1 meter second ^ -4 * [t(x)]^3

v(x) = 4*1 meter second^-4*[x^(0.25)*(1/12)^0.25*1 meter^-0.25 second ^1]^3

v(x) = 4 * (1/12)^0.75 * x^0.75 * 1 meter ^0.25 second ^ -1

KE(x) = 0.5 * 1 kilogram * [v(x)]^2

KE(x) = 8*(1/12)^1.5*x^1.5*1 kilogram meter ^0.5 second ^ -2

Dx(x^1.5) = 1.5*x^0.5

dKE(x)/dx = 12*(1/12)^1.5*x^0.5*1 kilogram meter ^0.5 second ^ -2

f(x) = dKE(x)/dx

int[x, f(x)] = KE(x) + K1

x(t) = 12*t^4* 1 meter second ^ -4

x(t = 1 second) = 12 meter second ^ -4 * (1 second)^4 = 12 meter

KE(x) = 8*(1/12)^1.5*x^1.5*1 kilogram meter ^0.5 second ^ -2

(1/12)^1.5*12^1.5=1

KE(x = 12 meter) = 8*1 Kilogram meter ^2 second ^ -2

v(t) = 4*t^3 * 1 meter second ^ -4

KE(t = 1 second) = 0.5 * 1 kilogram * [4 meter second ^ -1] ^ 2

KE(t = 1 second) = 8 kilogram meter ^ 2 second ^ -2

average force with respect to displacement from 0 meter to 12 meter =

=(int[x, f(x = 12 meter)] - int[x, f(x = 0)])/(12 meter - 0 meter) =

= 8 kilogram meter ^ 2 second ^ -2 / 12 meter =

= (2/3) * 1 kilogram meter second ^ -2

v(t) = 4 * t^3 * 1 meter second ^ -4

P(t) = 1 Kilogram * v(t)

P(t) = 4 * t^3 * 1 Kilogram meter second ^ -4

P(t = 1 second) = 4 * (1 second)^3 * 1 Kilogram meter second ^ -4

P(t = 1 second)  = 4 kilogram meter second ^ -1

f(t) = dP(t)/dt = 12 * t^2 * 1 Kilogram meter second ^ -4

int[t, P(t)] = P(t) + K2

average force with respect to time from 0 second to 1 second =

=(int[t, f(t= 1 second)] - int[t, f(t = 0)])/(1 second - 0 second) =

=  4 kilogram meter second ^ -1 / 1 second =

=  4 kilogram meter second ^ -2

f(x) = 12*(1/12)^1.5*x^0.5*1 kilogram meter ^0.5 second ^ -2

f(x=12meter)=12*(1/12)^1.5*(12meter)^0.5 *1kilogram meter ^0.5 second ^ -2

12*(1/12)^0.5*12^0.5=12

f(x = 12meter) = 12 kilogram meter ^ 2 second ^ -2

f(t) = dP(t)/dt = 12 * t^2 * 1 Kilogram meter second ^ -4

f(t = 1 second) = 12 kilogram meter second ^ -2

In this example the instantaneous force as a function of time is equal to the instantaneous force as a function of displacement at the displacement value that would occur for a given time value.  But the average force as a function of time is not the same as the average force as a function of displacement for the displacement interval that matches with the time interval.

Copyright Carl Janssen 2021 December 16