Given $s=-16t^2+192t+144$, what is the velocity when $s=0$?

This is part of a larger optimization problem which I solved, except for this last part. The critical point occurs at $t=6$, so after $t=6$ the position function is decreasing since the slope ($s^{\prime}$) is negative.

To find the velocity when $s=0$, I tried $s=-16t^2+192t+144=0$, and quadratic formula gives the roots, which are $t=-12.71$ and $t=0.71$. Shouldn't we use the positive root, $t=0.71$, since $t$ must be positive, because $t$ is time and can't be negative?

The text answer says $-214.72$ but that comes from using the negative root, and I don't understand why negative time $t$ is used. Is $-214.72$ incorrect?


  • $\begingroup$ Don't you just evaluate the derivative at the time for which $s(t)=0$, which occurs twice since there are two real solutions for it. A plot of $s$ is here. You can see it hits $0$ twice. Evaluating $s'(t)$ at these times gives me a speed of 214.66 for one crossing and -214.66 for the other. $\endgroup$ Jan 24 '14 at 6:48

Multiply $-1$ on both sides of the equation and substitute $s=0$ to get $16t^2-192t-144=0$. Then, solve with result from quadratic variables. Let $a=16$, $b=-192$, $c=-144$ and solve with $$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$

  • $\begingroup$ Ok, but there will be two roots, which one will I use? thanks $\endgroup$
    – Emi Matro
    Jan 24 '14 at 5:27
  • 1
    $\begingroup$ $t$ must be positive. $\endgroup$
    – kmitov
    Jan 24 '14 at 5:29
  • $\begingroup$ @kmitov There is nothing in the problem statement that says $t$ must be positive. $\endgroup$ Jan 24 '14 at 6:57
  • $\begingroup$ @AnonSubmitter85 $t$ represents time, I forgot to mention it. So it must be positive $\endgroup$
    – Emi Matro
    Jan 24 '14 at 7:14
  • $\begingroup$ @bryansis2010 $-214.72$ comes from using the negative $t$ root value, $-12.71$, why is the negative root used here if $t$ must be positive since it is time? thanks. +1 $\endgroup$
    – Emi Matro
    Jan 24 '14 at 7:29




$-214,6625258 = -32*12,70820393+192$

  • $\begingroup$ the two roots are $t=-12.71$ and $t=0.71$, why is the negative root used to give an answer of $-214.72$? is that a mistake? Shouldn't we use $t=0.71$ instead of $t=-12.71$? $\endgroup$
    – Emi Matro
    Jan 24 '14 at 7:32
  • $\begingroup$ Dear sir, I used the positive root. When one the numerator and denominator have the same signs, the quotient im positive. $\endgroup$
    – kmitov
    Jan 24 '14 at 7:46

well Dear kmitov if we solve the quadratic equation we get to 2 roots 12.708 or -0.708

but this answer is not the answer quoted by user 437158 can you help me on this

further velocity and s has a realtion

ds/dt (differentail)

if we find velocity by differentiating

it is idependent of variable s=0

and ds/dt = -32t+192

if we put velocity v = o then we get the time when body was supposed to be staionery

at t= 192/32=0.1875 second , the body was sationery but where it was from its origin s=o

it means that we should find instantous velocity when t= 0.1875

so we may say that instanous velocity was vt= 192-6 units when s=0

thanks Stormer wiki ghubaroo@gmail.com

  • $\begingroup$ $s(t)=a t^2+ v_0 t+ s_0$ $\endgroup$
    – kmitov
    Jan 24 '14 at 6:06

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.