Question

An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the
distance travelled before takeoff.​

Answer 1

Answer:

Question: An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the

distance travelled before takeoff.

Explanation:

Answer:

Initial speed of airplane u=0 m/s

Acceleration of the plane a=3.2 m/s

2

and time taken t=3.28 s

Using formula, distance covered S=ut+

2

1

at

2

⇒ S=0(32.8)+0.5(3.2)(32.8)

2

=1721.3∼1720m

Answer 2

Given :-

• Initial Velocity = 0 m/s
• Time = 32.8 seconds
• Acceleration = 3.20 m/s²

To Find :-

• Distance travelled = ?

Answer :-

• Distance travelled = 1721.344 m

Explaination :-

By using second equation of motion we get :

• s = ut + ½ at²

Where, s is Distance , u is Initial Velocity, a is Acceleration and t is time taken.

Now, subsitute the given values in second equation of motion :

➺ s = 0 × 32.8 + ½ × 3.20 × 32.8 × 32.8

➺ s = ½ × 3.20 × 32.8 × 32.8

➺ s = 1.6 × 32.8 × 32.8

➺ s = 1.6 × 1075.84

➺ s = 1721.344 m

Hence,

• Distance travelled = 1721.344 m

Some other Formulas :

• Speed = d/t

• F = m. a

• a (bar) = v - v_0/t = Δv/Δt

• w = F x d

• P = W/Δt

• V = IR

• PE = mgh

• p = mv