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# 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?

### 2 Answers

- 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