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

Answer:

The distance is 300 m.

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

$s= ut+\dfrac{1}{2}at62$

Put the value in the equation

$1200=0+\dfrac{1}{2}\times a\times(30)^2$

$a=\dfrac{2\times1200}{30\times30}$

$a=2.67\ m/s^2$

We need to calculate the distance

Using equation of motion

$v^2=u^2+2as$

Put the value in the equation

$40^2=0+2\times2.67\times s$

$s=\dfrac{40^2}{2\times2.67}$

$s=299.62\approx 300\ m$

Hence, The distance is 300 m.