S= a+ vt-16t squared, how to look for t?

Answer 1

Use the quadratic formula.

See:

http://cheeser1.slyip.com/interspace/quadt.gif

Answer 2

The key to that, is clever use of parentheses. If you want to denote squared by typing, use ^, so x squared would be x^2

S=a+(vt-16t^2)

To begin the isolation of t, distribute it out of the parentheses:

S=a+t(v-16t)

This works because the expression t(v-16t) is just a different way of saying (vt-16t^2)

I hope this helps you!

Answer 3

S = a + vt - 16t^2

subtract a for both sides
S - a = vt - 16t^2

divide both sides by -1
a - S = 16t^2 - vt

Divide 16 for both sides
(a - S)/16 = t^2 - (v/16)t

add (b/2)^2, or ( (v/16)/2)^2 for both sides
(a - S)/16 + ( (v/16) / 2)^2 = t^2 - (v/16)t + ( (v/16)/2)^2

(a - S)/16 + ( (v/32)^2 = t^2 - (v/16)t + ( (v/32)^2

factor
(a - S)/16 + ( (v/32)^2 = (t - v/32)^2

take square root for both sides
t - v/32 = +/- sqrt [ (a - S)/16 + ( (v/32)^2 ]

add v/32 for both sides
t = v/32 +/- sqrt [ (a - S)/16 + ( (v/32)^2 ]

Answer 4

The formula is incorrect. If you have a constant accleration of 32 ft/sec^2 body dropped at time=0 will, at time=t, have moved a distance = -16t^2 feet, with t in seconds. If the body had a velocity vo ft/sec at time=0, the formula will be....
vo*t -16 t^2 .