Speed at Takeoff and Average Acceleration

[Olivia Tarca]

Homework Statement

Starting from rest an airplane lifts off the ground after 40s and 1.8km. What is its speed at takeoff and its average acceleration during the takeoff run?

Homework Equations

V= distance/time
Acceleration= (Vf-Vi)/time

The Attempt at a Solution

The airplane travels 1.8km in 40 seconds. This would mean that when substituting into the formula V=d/t the velocity at takeoff is .045km/s or 45m/s. I then plugged in this value into the acceleration formula a= (45-0)/40s and got an answer of 1.125 m/s^2.
The answer key says that that answer should be 90m/s and then 2.25 m/s^s and I'm not sure why they are getting that answer?

 

[phinds]

The airplane travels 1.8km in 40 seconds. This would mean that when substituting into the formula V=d/t the velocity at takeoff is .045km/s or 45m/s.

Are you sure about that? Think about the actual mechanics of an airplane takeoff from rest at one end of the runway to liftoff at the other end.

 

[Olivia Tarca]

Are you sure about that? Think about the actual mechanics of an airplane takeoff from rest at one end of the runway to liftoff at the other end.

The airplane would start off slow and then reach a faster speed at takeoff which leads me to believe that the average velocity would be slower than its speed at takeoff, but I'm not sure how to go about calculating that and that doesn't correspond to anything I have calculated.

 

[phinds]

The airplane would start off slow and then reach a faster speed at takeoff which leads me to believe that the average velocity would be slower than its speed at takeoff, but I'm not sure how to go about calculating that and that doesn't correspond to anything I have calculated.

You are on the right track. What happens if you assume a smooth increase in speed from start to takeoff? You know the travel time and the travel distance.

 

[phasacs]

distance: x=vi+1/2at^2 is also a relevant equation, and since there is acceleration v=x/t (velocity=distance/time) is Not true

 

[TomHart]

distance: x=vi+1/2at^2 is also a relevant equation, and since there is acceleration v=x/t (velocity=distance/time) is Not true

Please double-check that equation; it isn't quite right.
And you are correct that v = x/t does not apply if there is acceleration.

Edit: Oh, I thought I was responding to the original poster. Oh well, I guess I'll stick with my response. :)

 

[haruspex]

v=x/t (velocity=distance/time) is Not true

It is true given the right definitions of variables. In that equation, v is the average velocity. The error was in misapplying the equation to a different v.

It might seem that I am just being pedantic, but I feel students can get confused if they think equations are right in some cases and wrong in others. The thing to stress is that an equation only means anything given the definitions of the variables it mentions. In this case, the student's notes could be emended to "v_avg = Δx/Δt".

 

[Zach S]

you need to use uniformly accelerated motion equations instead. If you don't know what these are I suggest reading through the beginning of your textbook.

the equations you will need are position equation and velocity equation.

 

[phasacs]

Please double-check that equation; it isn't quite right.

True, I forgot to multiply velocity with time: x=vi*t+1/2at^2