Question

An aeroplane accelerates down a ranway at 3.20 m/s² for 40 second untill it finally .lifts off the ground determine the distance traveled by the aeroplane on ranway​

Answer 1

Given : -

An aeroplane had accelerated down a runway at 3.2 m/s² for 40 seconds untill it finally lifts off from the ground .

Required to find : -

• Distance travelled by the aeroplane on the runway ?

Equations used : -

1) v = u + at

2) v² - u² = 2as

Here,

• v = Final velocity

• u = Initial velocity

• a = acceleration

• s = displacement

• t = time taken

Solution : -

An aeroplane had accelerated down a runway at 3.2 m/s² for 40 seconds untill it finally lifts off from the ground .

We need to find the distance travelled by the aeroplane on the runway ?

From the given information we can conclude that ;

Initial velocity of the plane ( u ) = 0 m/s

( Because, the planes will be at rest before starting )

Acceleration of the plane ( a ) = 3.2 m/s²

Time taken by the plane ( t ) = 40 seconds

Now,

Let's find the final velocity

Using the equation of motion ;

i.e. v = u + at

v = 0 m/s + 3.2 m/s² x 40 s

v = 0 m/s + 3.2 m/s² x 10 x 4 s

v = 0 m/s + 32 m/s² x 4 s

v = 0 m/s + 128 m/s²/s

v = 0 m/s + 128 m/s

v = 128 m/s

Hence,

Final velocity of the plane ( v ) = 128 m/s

Now,

Let's find the distance travelled by the plane .

Using the Equation of motion !

i.e. v² - u² = 2as

( 128 )² - ( 0 )² = 2 x 3.2 x s

16,384 - 0 = 6.4 x s

16,384 = 6.4s

6.4s = 16384

s = 16384/6.4

s = 2,560 meters

Hence,

Displacement of the plane = 2,560 meters

Therefore,

Distance travelled by the aeroplane on the runway is 2,560 meters

- : Additional Information : -

1st law of motion ;

In this universe , everybody continues to be in the state of rest or motion untill or unless some external force is applied on it .

2nd law of motion ;

The rate of change of momentum is directly proportional to the unbalanced force in the direction of force

3rd law of motion ;

For every action there is an equivalent opposite reaction .

Answer 2

Given; acceleration is 3.20 m/s², time is 40 sec and initial velocity is 0 m/s.

To find: distance traveled by the aeroplane on ranway.

Using the first equation of motion,

v = u + at

Substitute the values,

v = 0 + 3.20(40)

v = 0 + 128

v = 128

Now, using the second equation of motion,

s = ut + 1/2 at²

Substitute the values,

s = 0(40) + 1/2 × 3.20 × (40)²

s = 0 + 1.6 × 1600

s = 0 + 2560

s = 2560

Hence, the distance covered by the aeroplane on rranway is 2560 m.