Seemingly easy mechanics problem (constant acceleration)

In summary, the conversation discusses a problem involving a jet plane coming to a stop and the minimum time interval needed for this to occur. The person solving the problem attempted to use calculus to find the velocity function, but made a mistake by integrating the acceleration incorrectly. The correct formula for constant acceleration is v = u + at, and after rearranging, the person was able to solve the problem correctly.
  • #1
Feodalherren
605
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?
 
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  • #2
Check where that power of 2 came from.

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

1. What is constant acceleration?

Constant acceleration is the rate of change of velocity that remains the same throughout the motion of an object. This means that the object's velocity increases or decreases by the same amount in each unit of time.

2. How do you calculate acceleration?

Acceleration can be calculated by dividing the change in velocity (final velocity - initial velocity) by the time taken for that change to occur. The units of acceleration are typically meters per second squared (m/s²).

3. What is the difference between velocity and acceleration?

Velocity is the rate of change of an object's position, while acceleration is the rate of change of an object's velocity. In simpler terms, velocity describes an object's speed and direction, while acceleration describes how quickly that velocity is changing.

4. Can an object have constant acceleration if its velocity is changing?

Yes, an object can have constant acceleration even if its velocity is changing. This is because acceleration takes into account both the magnitude and direction of the change in velocity, and a change in either of these factors can result in a constant acceleration.

5. What are some real-world examples of constant acceleration?

Some examples of constant acceleration in everyday life include a car accelerating at a constant rate on a straight road, an object falling due to gravity, and a pendulum swinging back and forth.

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