Kinematics - How do I apply them to different problems?

[07triumphd675]

Kinematics -- How do I apply them to different problems?

Homework Statement

A Jet plane lands with a speed of 100 m/s and can accelerate at a maximum rate of -5.00 m/s^2 as it comes to a rest. A) From the instant the plane touches the runway, what is the min. time needed before it can come to a rest? B) Can this plane land on a small tropical island airport where the runway is 0.800 km long?

Homework Equations

Constant acceleration (Velocity as a function of time): Vf=Vo + at

Displacement as a function of time: X = Vo + t(1/2)(a)(t^2)

Velocity as a function of displacement: V^2 = Vo^2 + 2ax

The Attempt at a Solution

So I'm having a really tough time figuring out which equation to use where. I've deciphered what we're given.
Vf: 100 m/s
a = -5.00 m/s^2

The problem is asking in A) to find the time B) once we've found the time, we plug it back into one of the original equations (I would say the second one) to find the distance required for the plane to stop.

So, I take a stab at the first equation:

Vf = Vo + at

Vf = 100 m/s = Vo (I don't know the initial velocity) + (-5.00 m/s^2)(t)

So trying to solve for t: t= 105 m/s^3 - Vo

That doesn't make sense really. I keep getting hung up on these kinematic equations -- they keep coming back to haunt me in the next chapter when working on not only the x-axis but now the y-axis as well.

 

[Dick]

You are having some algebra problems as well. I would say Vo=100m/s (since that's the initial velocity when it touches down) and Vf=0m/s (since it stops). So 0m/s=100m/s+(-5m/s^2)*t. Try solving that for t. Be careful.

 

[edziura]

x = Vo + t(1/2)(a)(t^2) (1)

Two unknowns: x and t

V^2 = Vo^2 + 2ax (2)

The term on the left is also (Vf)^2; in this problem, the plane is slowing down an Vf is 0. solve this equation for x, and substitute the result into (1) and solve for t. then you can solve for x to answer the second part.

 

[07triumphd675]

Ahh okay. I got it now.

So solving for T using the first kinematics equation:

Vf = 0 m/s and Vo = 100 m/s

Vf = Vo + at

0 = 100m/s + (-5 m/s^2)(T)

-100 m/s = (-5 m/s^2)(T)

-100 m/s / -5 m/s^2 = T

T= 20 s

Using T then, plug it back into the second kinematics formula to find the distance required to land on the run way.

X = 100 m/s (20 s) + (1/2)(-5 m/s^2)(20^2)

Solving for x = 1000 m

Convert the 1000 m into km and you get 1 km required to land. Thus the answer is no since they're asking for .800 km.

That provides an awesome starting point. Thank-you so much.

 

[Dick]

Well done.