Grade 12 Calculus plane deceleration question

In summary, the position of an F-18 jet landing on a runway can be defined by the function s(t)=189t-t^7/3 meters after t seconds. To calculate the braking time, the derivative of this function is taken and set equal to zero. This results in a braking time of 0.92448 seconds. To find the deceleration at the instant the jet stops, the second derivative of the function is taken, which results in a deceleration of -84/9 m/s^2.
  • #1
speeddman
22
0

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
 
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  • #2
speeddman said:

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
@rock.freak667

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

well i don't know wouldn't taking the derivative help me find how long it takes to stop?

Ok what does the derivative represent ?
 
  • #5
the derivative would represent the deceleration wouldn't it? because the function represents the time.
 
  • #6
***distance/time
 
  • #7
speeddman said:
***distance/time

Which is velocity right? So if it is braking, what should the final velocity be?
 
  • #8
...0?
 
  • #9
speeddman said:
...0?

Right, so what do you get for t when you equate s'(t) to zero?
 
  • #10
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
speeddman said:

The Attempt at a Solution



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

speeddman said:
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
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
no it doesnt. 1.3333333333333root of 81 = 27s
 
  • #14
speeddman said:
no it doesnt. 1.3333333333333root of 81 = 27s

Yes 27s works better.

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

s''(t)=(-28/9)t^1/3
 
  • #16
speeddman said:
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
(-84/9) m/s^2

thank you so much by the way for everything!

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

Thanks for everything by the way!

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

1. How do you determine the plane's deceleration using calculus in grade 12?

In order to determine the plane's deceleration, you would need to use the formula for acceleration: a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time. By rearranging the formula to solve for a, you can find the deceleration of the plane.

2. What are the necessary variables needed to solve for the plane's deceleration?

The necessary variables needed to solve for the plane's deceleration include the initial and final velocities, as well as the time. These can be obtained from the given information in the problem, such as the distance traveled and the time it took to travel that distance.

3. Can calculus be used to solve for deceleration in other situations besides a plane?

Yes, calculus can be used to solve for deceleration in any situation where there is a change in velocity over time. This could include a car braking, a roller coaster slowing down, or even a person jumping off a diving board.

4. What other concepts in calculus are important to understand when solving for deceleration?

In order to solve for deceleration, you would need to have a good understanding of derivatives and integrals. Derivatives are used to find the rate of change of velocity, while integrals are used to find the total change in velocity over a certain time interval.

5. How can understanding deceleration in calculus be applied in real-world situations?

Understanding deceleration in calculus can be applied in real-world situations such as engineering, physics, and even sports. Engineers use calculus to design vehicles and structures that can safely decelerate, while physicists use it to study motion and forces. In sports, understanding deceleration can help athletes improve their performance by optimizing their movements and reducing the risk of injury.

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