Question

Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down the runway. A smaller plane with the same acceleration has a takeoff speed of 40 m/s. Starting from rest, after what distance will this smaller plane reach its takeoff speed? Knight, Randall D., (Professor Emeritus). College Physics (p. 64). Pearson Education. Kindle Edition.

Answer 1

The distance is 300 m.

Step-by-step explanation:

Given that,

Time = 30 s

Speed = 80 m/s

Distance = 1200 m

Speed of smaller plane = 40 m/s

We need to calculate the acceleration

Using equation of motion

`s= ut+(1)/(2)at62`

Put the value in the equation

`1200=0+(1)/(2)* a*(30)^2`

`a=(2*1200)/(30*30)`

`a=2.67\ m/s^2`

We need to calculate the distance

Using equation of motion

`v^2=u^2+2as`

Put the value in the equation

`40^2=0+2*2.67* s`

`s=(40^2)/(2*2.67)`

`s=299.62\approx 300\ m`

Hence, The distance is 300 m.