# Finding velocity in optimization problem

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?

Thanks.

• 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. 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 $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

• Ok, but there will be two roots, which one will I use? thanks Jan 24 '14 at 5:27
• $t$ must be positive. Jan 24 '14 at 5:29
• @kmitov There is nothing in the problem statement that says $t$ must be positive. Jan 24 '14 at 6:57
• @AnonSubmitter85 $t$ represents time, I forgot to mention it. So it must be positive Jan 24 '14 at 7:14
• @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 Jan 24 '14 at 7:29

D=11520

$107,3312629=\sqrt{11520}$

$12,70820393=(-96-107,3312629)/(-16)$

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

• 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 $t=-12.71$? Jan 24 '14 at 7:32
• Dear sir, I used the positive root. When one the numerator and denominator have the same signs, the quotient im positive. 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

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