Rearrange this s=vt-(1/2)at^2 and s=(1/2)at^2 +ut to make t the subject?
Im struggling a little bit with motion formulas and when I rearrange to make t subject I get different results with each formula.
Can somebody give me a tip on how I should use 'constant acceleration' motion formulas?
- billrussell42Lv 71 month agoFavorite Answer
of course you would get different results, the two equations are different.
s = vt – (1/2)at²
at² – 2vt + 2s = 0
quadratic equation:to solve ax² + bx + c = 0x = [–b ± √(b²–4ac)] / 2at = [2v ± √(4v²–8as)] / 2a
t = [v ± √(v²–2as)] / a
t = (v/a) ± (1/a)√(v²–2as)
s = (1/2)at² + ut
at² + 2ut – 2s = 0
x = [–b ± √(b²–4ac)] / 2at = [–2u ± √(4u²–8as)] / 2a
t = [–u ± √(u²–2as)] / a
t = –(u/a) ± (1/a)√(u²–2as)
correct equation is
Equations of motion (straight line, constant acc)
d = ½at² + v₀t + d₀
d is displacement
v₀ is initial velocity
d₀ is initial position
if you need to solve for t, best it to have numbers for the other terms.
- oldschoolLv 71 month ago
Let Vi = initial velocity and Vf = final velocity.
Vf = Vi + a*t so t = (Vf-Vi)/a
Distance s = (Vf+Vi)/2 *t = (Vf+Vi)/2 *(Vf-Vi)/a = (Vf²-Vi²)/2a = s
Vf² - Vi² = 2*a*s
s = ∫V(t)dt = ∫Vi +a*t dt = Vi*t + ½*a*t²