If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Class 9 Physics (India)

### Course: Class 9 Physics (India)>Unit 1

Lesson 4: Acceleration

# Airbus A380 take-off time

Figuring how long it takes an A380 to take off given a constant acceleration. Explore the physics of an Airbus A380 take-off with Khan Academy. Learn how to calculate the time it takes for the aircraft to reach takeoff velocity using principles of acceleration and velocity. This practical example brings the concepts of speed, time, and acceleration to life. Created by Sal Khan.

## Want to join the conversation?

• Quick question, does the weight of the Airbus A380 play a part in how fast the aircraft accelerates?
• The heavier the craft the more force is needed to reach a specific acceleration.
• In which lessons can I learn this way of doing maths, with equations, cancelling out things and so on?
• Type in 'dimensional analysis' in the search bar near the top of your page and it will bring you to the section in algebra 1 that deals with these types of conversions. Developing a solid understanding of this concept while in algebra will really help you in many subjects.
• This may be covered later, or just sloppy thinking on my part, but is there some sort of something beyond acceleration? like, meters/second/second/SECOND? that would be an increase in the rate of acceleration? or is that just change in acceleration and I'm being confusing and dramatic? I mean, mathematically obviously we can do this, but does it make sense in the real world?
• That is exponential acceleration. Not only is velocity increasing, but the acceleration that increases the velocity is increasing!
• If an aircraft is smaller will the speed be more faster? For ex. a Boeing 737/Airbus 320.
• The aerodynamic shape of the plane and its engines are more important to its speed than the size. For example, a Concorde can travel faster than a F-104 Nighthawk eve though it is nearly 10 times larger because it is sleek and light and has 4 Rolls-Royce engines powering it to supersonic speed whereas the Nighthawk is built for spying so it has heavier "cloaking" material and crooked sides that disperse RADAR waves so it cannot be detected thus increasing its drag. Size doesn't necessarily matter!
• Gee, still can't understand why there is square in the m/s^2 formula. if every second smth accelerates x m/s, why then there is not 2s or m/2s?
• velocity tells you how many meters you move per second: hence m/s
acceleration tells you how much the velocity (m/s) changes per second: hence (m/s)/s which is m/s^2!
• I don't understand why would you multiply 280 by 1/3600 and 1000/1
• For example, suppose we wish to convert 15.0 in. to centimetrse.
Because 1 in. is defined as exactly 2.54 cm, we find that
15.0 in. =15.0 in. *2.54 cm/1 in.
= 38.1 cm
where the ratio in parentheses is equal to 1. We express 1 as 2.54 cm/1 in. (rather than 1 in./2.54 cm) so that the unit “inch” in the denominator cancels with the unit
in the original quantity. The remaining unit is the centimeter, our desired result.
• Why velocity over acceleration gives us the time of take off? How do we know that?
• Because acceleration is how much velocity you gain per second. So, assuming the initial velocity is 0 and a constant acceleration, dividing the final velocity by the acceleration give you the time it took to reach this final velocity :
Vf = Vi + a*t
and because the initial velocity is 0 :
Vf = a*t
and because the acceleration is constant :
t = Vf/a
where :
Vi = initial velocity
Vf = final velocity
a = acceleration
• I was wondering, the 78 seconds Mr. Khan got is just an estimate of the total answer. So if you would want to know the exact answer, would you need more information or would you use a different equation?
• I believe you would need to use a different equation because the actual acceleration of the airplane is not constant, but you need a constant acceleration to use the equations shown in this video.
• Can we find the distance traveled by and object if the acceleration of the object is given.
I know that Acceleration = change in velocity / total time taken.
So, If a car accelerated from 5 m/s to 25 m/s in 10 seconds, how far will it travel ?
Acceleration = 20/10 = 2m/s^2
But from the acceleration how can we find the Distance ?
• basically its simple! I am in class seven still though but i can i can help you out in this!

The formula is "s=ut+1/2at^2"

I hope you can easily understand the cause,
Thank you.
(Forgive me for any mistakes)