Question

An aeroplane accelerates down a runaway at 3 ms-2 for 30 s until it finally lifts off the ground. Determine
the distance travelled before take off.​

Answer 1

Answer:

1.35 km (or) 1350 metres

Explanation:

Given :

• Acceleration of the aeroplane = a =3 m/s²

• Time the aeroplane accelerates for = t = 30 seconds

• Initial velocity = u = 0 m/s (As the plane starts from rest while take off)

To find :

• The distance travelled by the aeroplane before take off

Using the first equation of motion :

V=u+at

V=0+3×30

V=90 m/s

Now using the third equation of motion:

V²-u²=2as

90²-0²=2×3×s

8100-0=6×s

8100=6s

S = 8100/6

S = 1350 metres

The distance travelled by the aeroplane before takeoff is 1.35 km or 1350 metres

Answer 2

Answer:

Given,

the initial velocity = 0 m /s.

acceleration = 3.20 m / s^2

time = 32.8 s

According to laws of motion.

s = ut + 1/2 at ^2

s = 1/2 at²

s=1/2(3.20)(32.8)²

s= 1721.344 m

the distance traveled before takeoff is 1731.3m