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Grade 12 Calculus plane deceleration question

  • Thread starter speeddman
  • Start date
  • #1
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Homework Statement



When an F-18 jet lands on the runway, it's position is defined by s(t)=189t-t^7/3 meters after t seconds

calculate the braking time

calculate the deceleration when the jet stops ( i think she means immediately before?)

Homework Equations



derivative

The Attempt at a Solution



s'(t)=189-7/3t^4/3

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
rock.freak667
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The Attempt at a Solution



s'(t)=189-7/3t^4/3
What about the derivative, you want to find a t for which s'(t) is what value?
 
  • #3
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@rock.freak667

well i dunno wouldn't taking the derivative help me find how long it takes to stop?
 
  • #4
rock.freak667
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@rock.freak667

well i dunno wouldn't taking the derivative help me find how long it takes to stop?
Ok what does the derivative represent ?
 
  • #5
22
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the derivative would represent the deceleration wouldn't it? because the function represents the time.
 
  • #6
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***distance/time
 
  • #7
rock.freak667
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***distance/time
Which is velocity right? So if it is braking, what should the final velocity be?
 
  • #9
rock.freak667
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...0?
Right, so what do you get for t when you equate s'(t) to zero?
 
  • #10
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81=t^4/3

so t=0.92448s

so it takes 0.92448s for the plane to stop.

now for to find the deceleration at an instantaneous pnt i have to take second derivative?
 
  • #11
rock.freak667
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The Attempt at a Solution



s'(t)=189-7/3t^4/3
81=t^4/3

so t=0.92448s

so it takes 0.92448s for the plane to stop.

now for to find the deceleration at an instantaneous pnt i have to take second derivative?

You have two differing statements. Check it over.
 
  • #12
22
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well s'(t)=189-7/3t^4/3

0=189-7/3t^4/3

-189/(-7/3)=t^4/3

81=t^4/3

4/3root of 81= t

t=0.92448s

does that make sense?
 
  • #13
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no it doesnt. 1.3333333333333root of 81 = 27s
 
  • #14
rock.freak667
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no it doesnt. 1.3333333333333root of 81 = 27s
Yes 27s works better.

Now how would you get the acceleration from s'(t)?
 
  • #15
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second derivative?

s''(t)=(-28/9)t^1/3
 
  • #16
rock.freak667
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second derivative?

s''(t)=(-28/9)t^1/3
and if it takes 27 seconds to stop, what is the deceleration at that instant?
 
  • #17
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(-84/9) m/s^2

thank you so much by the way for everything!

is that right?
 
  • #18
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(-84/9) m/s^2

Thanks for everything by the way!

is that right?
 
  • #19
rock.freak667
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Yes that is correct.
 

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