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Seemingly easy mechanics problem (constant acceleration)

  • #1
599
6

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?
 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
Check where that power of 2 came from.

You should know a family of constant acceleration formulae, including
v = u + at
 
  • #3
599
6
I made it up as I took the integral. I'm not sure what you're getting at here. What do you mean?
 
  • #4
10
0
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.
 
  • #5
33
3
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...
 
  • #6
599
6
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.
 
  • #7
33
3
Yes when you integrate a = dv/dt = -5 !!! you get v = -5t + v0 (=100) not v= -(5/2)t +100. ;)
 
  • #8
Doc Al
Mentor
44,905
1,169
I integrated a= -5t not -2.5t^2.
The acceleration is -5, not -5t. It's constant.
 
  • Like
Likes 1 person
  • #9
599
6
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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