Question

An air-plane accelerates down a runway at 3.20 m/s^2 for 32.8 s until is finally lifts off the ground. Determine the distance travelled before take-off.

Answer 1

Answer:

1720 m

Explanation:

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

Answer:

The air plane covers a distance of 1721.344 before it lifts off the ground.

Explanation:

Given,

Acceleration (a) = 3.2 m/s^2

Time (t) = 32.8 s

Initial Velocity (u) = 0 m/s

S = ut+1/2at^2

= 0.5*3.2*32.8*32.8 m

= 1721.344 m

Hope this helps.