F = mdt⁻²
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You're close, but not quite.
By Newton's second law:
F=md/t²
ma=md/t²
a=d/t²
this equation will be a little easier to check. consider the following kinematics equation (known to be true):
d=v1(t)+at²/2
In this case, d is the distance traveled, v1 is the original speed of the object, t is the time elapsed, and a is the acceleration. Note that for your formula above, we are assuming v1=0. So:
d=0+at²/2
d=at²/2
2d=at²
a=2d/t²
So now we have an expression to substitute in for a, so lets multiply both sides of the equation by m again.
ma=2md/t²
F=2md/t²
and there you have it, you were certainly close, just missing the factor of 2.
Cheers.
ps. elfmotat, he clearly stated what the elements were
By Newton's second law:
F=md/t²
ma=md/t²
a=d/t²
this equation will be a little easier to check. consider the following kinematics equation (known to be true):
d=v1(t)+at²/2
In this case, d is the distance traveled, v1 is the original speed of the object, t is the time elapsed, and a is the acceleration. Note that for your formula above, we are assuming v1=0. So:
d=0+at²/2
d=at²/2
2d=at²
a=2d/t²
So now we have an expression to substitute in for a, so lets multiply both sides of the equation by m again.
ma=2md/t²
F=2md/t²
and there you have it, you were certainly close, just missing the factor of 2.
Cheers.
ps. elfmotat, he clearly stated what the elements were
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Nope. F = ma
a = d''(t) = second derivative of displacement function
a is not always d / t^2. You can have zero acceleration, and still travel.
Just use the equations of motion to find a:
∆x = u*t + (1/2) at^2, where u is initial velocity
a = (v - u) / t, where v is final velocity, u is initial velocity
And the others, which are given in the link in my source
EDIT:
F = ma
a = (∆x - ut) / t^2
=> F = m(∆x - ut) / t^2 = m∆x / t^2 - u / t
∆x represents your displacement. u represents initial velocity.
If u = 0, then F = d / t^2. Otherwise, it's not so simple. You can't assume that initial velocity is zero.
a = d''(t) = second derivative of displacement function
a is not always d / t^2. You can have zero acceleration, and still travel.
Just use the equations of motion to find a:
∆x = u*t + (1/2) at^2, where u is initial velocity
a = (v - u) / t, where v is final velocity, u is initial velocity
And the others, which are given in the link in my source
EDIT:
F = ma
a = (∆x - ut) / t^2
=> F = m(∆x - ut) / t^2 = m∆x / t^2 - u / t
∆x represents your displacement. u represents initial velocity.
If u = 0, then F = d / t^2. Otherwise, it's not so simple. You can't assume that initial velocity is zero.
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It's important to realize that d is not equal to at². If you accelerate a particle from rest for t seconds, then the distance it will have traveled will be d = at² / 2. The one half factor is important. Your equation should be F = 2md / t².
@ ☼¿☼: I started writing my post before he had added the additional details.
@ ☼¿☼: I started writing my post before he had added the additional details.
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dunno, but when i am looking at equations it sometimes helps me to see if the units on both sides of the equals sign are the same (if they cancel out to be the same thing)
then the equation is usually correct
then the equation is usually correct
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If F=ma, and a=d/t^2, then it should be, F=m(d/t^2). (a=acceleration)
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well form my gr 10 physics knowledge F = ma. m being mass and a being acceleration. but idk how old u r so im guessing it gets more advanced than that...