anonymous
  • anonymous
Find the instantaneous velocity and speed at time 5 secs when v(t) = t2 – 9t + 18 and s(0) = 1
Mathematics
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anonymous
  • anonymous
Find the instantaneous velocity and speed at time 5 secs when v(t) = t2 – 9t + 18 and s(0) = 1
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
i figured i would just ave to find the acceleration (which is just the derivative of v(t)) and then plug five into that but i did that and got it wrong
Astrophysics
  • Astrophysics
Why would you find acceleration when it's asking for velocity, just plug in 5 for your velocity.
anonymous
  • anonymous
its asking for the instantaneous velocity doesn't that mean the tangent?

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Astrophysics
  • Astrophysics
Are you sure it's not suppose to be s(t)
anonymous
  • anonymous
no i am given v(t)
anonymous
  • anonymous
but i am also given s(0) so i can just find s(t)
Astrophysics
  • Astrophysics
You're given the instantaneous velocity
Astrophysics
  • Astrophysics
Just find v(5)
anonymous
  • anonymous
okay so then v(5) = -2
Astrophysics
  • Astrophysics
Yes
anonymous
  • anonymous
okay and for the speed i just find s(t) and plug five into that right?
Astrophysics
  • Astrophysics
s(t) is the position
Astrophysics
  • Astrophysics
speed is just scalar
anonymous
  • anonymous
okay so then speet(t)=2 ?
Astrophysics
  • Astrophysics
Yes, speed is just the magnitude or the velocity
anonymous
  • anonymous
okay lovely ! thank you very much!
Astrophysics
  • Astrophysics
Yw :)
Astrophysics
  • Astrophysics
We could find the position to s(t), by integrating, \[v(t) = t^2-9t+18 \implies \frac{ ds }{ dt } = t^2-9t+18\]\[\int\limits ds = \int\limits (t^2-9t+18) dt = s(t) = \frac{ t^3 }{ 3 }-\frac{ 9t^2 }{ 2 }+18t+C \]

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