Hi all,
Supposing that s' is the new position, s the old one, t the time started, t' the new time, Δt the time difference, (approaches zero) and v the velocity, or the derivative of the position with respect to time: s' = s + vt. So far so good, I'll work out the derivative first:
ds/dt = vt + vΔt - vt/vΔt
= v.
That shows that the derivative of ds/dt is v. However, most calculus equations are given as s = ƒ(t) which can be something like s = 4t^2 + 2t + 20. How is this possible? I know that the equation of a straight line can be written as : y = mx + c which in this case would be s' = vt + s. So how is it possible, without knowing the velocity, to find an equation such as the s = ƒ(t)? one here? Besides doesn't this equation hold only for a straight line? So would it apply throughout an entire graph if the velocity, that is the slope changes? And if so, why?
Any help would be greatly appreciated, thanks
Supposing that s' is the new position, s the old one, t the time started, t' the new time, Δt the time difference, (approaches zero) and v the velocity, or the derivative of the position with respect to time: s' = s + vt. So far so good, I'll work out the derivative first:
ds/dt = vt + vΔt - vt/vΔt
= v.
That shows that the derivative of ds/dt is v. However, most calculus equations are given as s = ƒ(t) which can be something like s = 4t^2 + 2t + 20. How is this possible? I know that the equation of a straight line can be written as : y = mx + c which in this case would be s' = vt + s. So how is it possible, without knowing the velocity, to find an equation such as the s = ƒ(t)? one here? Besides doesn't this equation hold only for a straight line? So would it apply throughout an entire graph if the velocity, that is the slope changes? And if so, why?
Any help would be greatly appreciated, thanks
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