How do you solve d=v0t+1/2at^2

Question

I have this on a math packet I have to finish today but idk how to do it. Please show the work.

Answer 1

d(t) = (a/2)t^2 + vot

where d= distance as a fuction of time t

a= acceleration, vo = initial velocity

d'(t) = v(t) = at +vo = velocity at time t

d"(t) = v'(t) = a(t) = a = a constant, independent of time t

for falling objects, a=g = -32 feet per second per second

or in the metric system a=g=-9.8 meters per second per second

or maybe the problem was to solve for vo, a, or t in terms of d?

solve for vo

d=at^2/2 +vot

vot = d-at^2/2

vo = (d-at^2/2)t

solve for t

at^2/2 = d-vot

t^2 +vot -d = 0

use the quadratic formula

t = -vo/2 +/- (1/2)sqr(vo^2+4d)

solve for a

at^2 = d-vot

a = (d-vot)/t^2