Basic Physics (Horizontal Motion)

Question

Please I need help with this one, provide an in-depth step by step with complete solution and final answer.

An airplane at rest accelerates on a runway at 5.25 m/s^2 for 30 seconds until it finally lifts off the ground. Determine the distance travelled by the airplane before takeoff. What is the velocity of the airplane at 25 seconds?

Thank you so much, It's the 2nd day of me learning Physics. God Bless!

Answer 1

Hi Renan C.,

In Physics, you need to understand what the equations of motion mean, physically, and be able to apply them. It would be best to memorize them, too, it will save you a lot of time. Another thing: get used to reading negative exponents on units, as "per (whatever) (optional exponent use)" -- so s^-2 is "per second squared".

So. a = acceleration = 5.25 m s^-2

t = time interval under consideration = 30 s

v = velocity, but you may treat it as speed, here

_{v(liftoff) = a*t}

v = 5.25*30 = 157.5 m s^-1

The problem didn't ASK for that, but it is convenient for plugging in to another convenient formula:

v_(f) ^2 = v_(i) ^2 + 2 a*d where v_(f) = final velocity, v_(i) = initial velocity

So (157.5)^2 = (0)^2 + 2*5.25*d (I've dropped the units for speed of typing here; in an actual problem you should always carry them, they are your methodology check!).

d = 2362.5 m

Let's also check by another equation for distance and acceleration:

d = (1/2) a t^2

d = 0.5*5.25*(30)^2 = 2362.5 m

And also v₍₂₅₎ = at = 5.25*25 = 131.25 m s^-1

By the way, that's quite a long distance for liftoff; commercial runways range from 2400 m to 3900 m. Which goes to show you why aircraft always take off and land against a significant wind -- their lift is better at the same ground speed, so they have better safety margins.

-- Cheers, -- Mr. d.

Answer 2

Solution:

The plane starts from rest, therefore initial velocity V_o = 0 , acceleration a = 5.25 m/s² and time T = 30 sec till lift off, so you need the travelled distance

The distance is expressed in this formula X = 1/2 a t² + V_o t + X_o where X_o is the initial distance = 0

Substituting X = 1/2 a t² + (0) ( V_o) + 0 , and X = 1/2 a t² , X = 1/2( 5.25 m/s²)( 30s)² = 2362.5 m

After 25 s the velocity of the plane is V = dX / dt = 1/2( 2) a t^2-1 ( 1) + V_o( 1) + 0

V = a t + V_o = at + 0 = at = 5.25 m/s² ( 25 s) = 131.25 m/s = 131.25 m/s

or (131.25 m/s )( km/1000m) ( 3600 s/ hr) = 472.5 km/hr