Seemingly easy mechanics problem (constant acceleration)

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SUMMARY

The discussion revolves around solving a mechanics problem involving a jet plane landing with an initial speed of 100 m/s and a maximum deceleration of 5.00 m/s². The participant attempted to derive the velocity function using calculus but incorrectly integrated the acceleration, leading to an erroneous formula. The correct approach utilizes the kinematic equation v = u + at, which confirms that the minimum time required for the plane to come to rest is 20 seconds. The key takeaway is the importance of applying established kinematic equations in constant acceleration scenarios.

PREREQUISITES
  • Understanding of kinematic equations, specifically v = u + at
  • Basic calculus, particularly integration techniques
  • Knowledge of constant acceleration concepts
  • Familiarity with mechanics problems involving motion
NEXT STEPS
  • Study the derivation and application of kinematic equations in physics
  • Practice solving problems involving constant acceleration using both calculus and kinematic formulas
  • Explore the relationship between acceleration, velocity, and time in motion scenarios
  • Review integration techniques in calculus, focusing on applications in physics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and motion, as well as educators seeking to clarify the application of kinematic equations in real-world scenarios.

Feodalherren
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Homework Statement


A jet plane comes in for a landing with a speed of
100 m/s, and its acceleration can have a maximum magnitude
of 5.00 m/s^2 as it comes to rest. (a) From the
instant the plane touches the runway, what is the minimum
time interval needed before it can come to rest?


Homework Equations


Kinematics I suppose, but I'm trying to solve it with just calculus.


The Attempt at a Solution



So my idea was that if the acceleration is constant at -5m/s^2 then I should be able to take the integral of this and get the velocity function. So what I end up with is

v(t) = - 5/2 (t^2) + 100

Solve for v(t) = 0 etc.

But the answer is wrong. It's supposed to be 20s.

Where is my logic flawed?
 
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Check where that power of 2 came from.

You should know a family of constant acceleration formulae, including
v = u + at
 
I made it up as I took the integral. I'm not sure what you're getting at here. What do you mean?
 
There are a few equations in mechanics that are linear and deal with constant acceleration.

V = u + at

is one of them
Where

U is initial velocity
V is final velocity
a is acceleration
t is time

If you rearrange the formula, you can work out the time taken.
 
With just calculus you just got wrong the formula of v. You have a= -2.5 . When you integrate that you have? not t^2 obviously...
 
That's the thing, I wasn't using the kinematic equations. It seems to me that I should be able to solve this using just calculus. My reasoning was that you can solve "gravity problems" that way. If you integrate -9.8t you get the velocity function for dropping something, -4.9t^2, correct?
Then why doesn't this work?

I integrated a= -5t not -2.5t^2.
 
Yes when you integrate a = dv/dt = -5 ! you get v = -5t + v0 (=100) not v= -(5/2)t +100. ;)
 
Feodalherren said:
I integrated a= -5t not -2.5t^2.
The acceleration is -5, not -5t. It's constant.
 
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Ohhh my gosh! Of course! Thank you so much Doc Al, you saved my day. That makes complete sense.

I lift my hat for you Sir!
 

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