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.

# Acceleration of aircraft carrier take-off

Using what we know about take-off velocity and runway length to determine acceleration. Created by Sal Khan.

## Want to join the conversation?

• Wouldn't the initial velocity change with how fast the ship is traveling? •   It depends on your point of reference. In this problem, the velocity is being measured relative to the starting point at the catapult (on the ship's deck). The catapult is point 0 (the initial reference position). During prelaunch, the F-18 is stationary at the catapult... not moving relative to its initial position (i.e. not rolling on the runway), hence its initial velocity is considered 0.

This example is somewhat simplified, but yes, the aircraft is moving with the ship, so it is moving relative to a fixed point on the globe. There are other factors for an aircraft taking off as well... most importantly, wind speed. Wind speed relative to the aircraft affects lift. Normally (operations allowing) a carrier will turn into the wind for the benefit of lift. For example, the ship could be going 25 knots against a 15 knot wind, effectively generating 40 knots of headwind. Add in the speed of the aircraft from catapult assist (and jet engines) through the wind, and you have some good lift, really fast. Once the aircraft no longer has catapult assist, it relies on its own power from the jet engines pushing against air (like in the prior video talking about rockets). Back to the relative position concept, an aircraft can move at a fixed airspeed (speed relative to air/wind), but relative to the ground, it will travel faster with the wind and slower against a headwind. With the heavier F-14s especially, you'll often notice a bit of a dip in the initial ascent off the deck... then an afterburner kick to get airspeed up.
• instead of all that can we do 80m/(36m/s) would that be the same thing because i divided the average velocity witch is 36m/s by the distance would that be more common sense? •  Yes. I did the same thing. I found the time first by using the fact that the time= (displacement) / (avg velocity)= 2.2s. Then, I used the time to find acceleration from the fact that change in velocity= acceleration x time. Therefore, acceleration= delta v/t or (72m/s)/(2.2s)
• Im just trying to get a better understanding of acceleration. Sal said that while going 33m/s squared- after 1 second hes going 33m/s, after 2 seconds hes going 66m/s, how fas is he going after the third second? is it 99m/s? or is it 132m/s?? •  99m/s. The rate of acceleration (how much velocity is added each second) is constant. So each second his velocity will be greater by 33 m/s regardless of his current velocity
• I got more or less his answers, but without using squares:

Displacement = Average Velocity * time; I found time to be 2.22 seconds
(Displacement is the runway's length, 80 meters and the Average Velocity is about 36 m/s)

Acceleration = Velocity / Time so I got an acceleration of 32.5 meters/second^2
(Velocity is 72 m/s and Change in time is 2.22 seconds, as found above)

I did get the same answers, but I didn't use the same methods... should I be worried? Which method is preferred? • I found a simpler way of finding the acceleration.
Given-
Displacement = Runway length = 80m
Initial velocity = 0
Final velocity = Take off speed = 260km/hr = 72m/s
Time = unknown

The solution-
1st part: Finding the time.
Vavg = s/t
t = s / Vavg
t = s / ((Vf+Vi)/2)
t = s / ((72m/s+0m/s)/2)
t = s / (72/2)
t = 80/36
t= 2.2 sec
2nd part: Finding the acceleration
a = Δv/t
a = ((72-0)m/s)/2.2sec
a = 72m/s / 2.2sec
a = 32.7m/(s*s)
So, isn't this method right? • Thinking of aerofoils, would a tailwind during flight increase the minimum speed needed to maintain lift? Let me think out loud: the wind would also reduce the needed energy to maintain the momentum vs air resistance and if the energy exerted is constant, it would accelerate the plane. I guess the answer depends on if the tailwind and force exerted cancel out? • would the movement of the ship on the sea make the plane have a different take off each time • Does Vavg = x/t, equal to Vavg = (vf+vi)/2? • Only under constant acceleration conditions. This is because v_avg = total disp/total time which can be written as (if initial velocity is u and accn a, time interval t) [ut+(1/2)at^2] / [t] which is equal to u+1/2at and if final velocity is v, then v = u+at. Therefore, at = v-u, Substituting, v_avg=u+(v-u)/2 = (v+u)/2.

Hence, both the average velocities are one and the same.  