Question

Two particles are moving on an axis. The positions for the two particles when t is less than or equal to 0 are: s1(t)=t^2-5t-50 and s2(t)= t^3/3-4t^2-9t. When are the particles traveling at the same velocity? (I already have the first and second derivative for both functions)

Answer 1

Well if they travel at the same velocity then the first derivative of both should be the same :)

Answer 2

Set the first derivatives equal to each other, then solve for t.

Answer 3

i thought i would have to set the derivatives equal to eachother and solve for t which what i began to do until i got stuck

Answer 4

I am sorry I was wrong.

Answer 5

What you have to do is find the first derivative of both and solve for t. I'll show you the maths.

Answer 6

Show us where you got stuck, @jstephanie

\[\large v _{1}(t)=2t-5 \]
\[\large v _{2}(t) = t^2 -8t -9\]

Equate them to find where they're equal: \[\large 2t-5 = t^2 -8t -9\]

Answer 7

\[\frac{ d }{ dt }S _{1}=\frac{ d }{ dt }S _{2}=v _{1}=v _{2}=2t-5=t ^{2}-8t-9 \rightarrow 0=t ^{2}-10t-4\] Then you apply the quadratic formula and:\[t=\frac{ 10\pm \sqrt{10^{2}-4(-4)} }{ 2 }=\frac{ 10\pm \sqrt{116} }{ 2 }\]

Answer 8

I didnt set it up like that thats probably where I went wrong @ivancsc1996 I did this:
\[2t-5=t ^{2}-8t-9\rightarrow 2t= t ^{2}-8t-4\rightarrow 0= t ^{2}-6t-4\] then i got stuck because i couldnt get it to be a binomial

Answer 9

@agent0smith

Answer 10

You made some mistakes, @jstephanie when you subtracted 2t from both sides:

\[\large 2t-5 = t^2 -8t -9 \] add 5 to both sides

\[\large 2t = t^2 -8t -4\] subtract 2t from both sides

\[\large 0 = t^2 -10t -4\]

Now use the quadratic formula as @ivancsc1996 showed (it won't factor)

Answer 11

ohh I see... I added instead of subtracting... Okay thank you