1
$\begingroup$

I'm using an intermediate algebra textbook and it had this problem:

"Solve the formula $$h = 48t + \frac{1}{2}a t^2$$ for $t$." The answer they displayed was: $$a = \frac{2h-96t}{t^2}$$

Can anyone tell me how this answers the question "solve for $t$", and what would be the right way to solve for $t$?

$\endgroup$
  • 1
    $\begingroup$ That is solving for a. $\endgroup$ – Ian May 18 '19 at 3:53
  • $\begingroup$ Maybe they meant solve for a. $\endgroup$ – NoLand'sMan May 18 '19 at 3:54
  • $\begingroup$ So, what would the answer for solving for t? $\endgroup$ – Joe May 18 '19 at 4:05
1
$\begingroup$

If you want to solve for $a$, you want to isolate $a$ to get it by itself.

So first, I'd subtract $48t$ from both sides.

$$h - 48t = \frac{1}{2}at^2$$

Then you can multiply both sides by 2:

$$ 2(h-48t) = at^2$$

and finally dividing both sides by $t^2$

$$a =\frac{2h-96t}{t^2}$$

Edit: to solve for $t$, you can use the quadratic formula. For $Ax^2+Bx+C = 0$, we have that

$$x = \frac{-B \pm \sqrt{B^2-4AC}}{2A}$$

So if we have $\frac{1}{2}at^2+48t-h = 0$, we can plug into the equation with $A = \frac{1}{2}a$, $B = 48, C=-h$:

$$t = \frac{-48 \pm \sqrt{48^2 - 4(\frac{1}{2}a)(-h)}}{2(\frac{1}{2}a)}$$

$\endgroup$
  • $\begingroup$ What would be the answer for solving for t? $\endgroup$ – Joe May 18 '19 at 4:05
  • $\begingroup$ @user65141 see my edit $\endgroup$ – rb612 May 18 '19 at 4:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.