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?

Answer 1

300 m

Step-by-step explanation:

The train accelerate from the rest so u = 0 m/sec

Final speed that is v = 80 m/sec

Time t = 30 sec

The distance traveled by first plane = 1200 m

We know the equation of motion
`S=ut+(1)/(2)at^2` where s is distance a is acceleration and u is initial velocity

Using this equation for first plane
`1200=0* 30+(1)/(2)a30^2`

`a=2.67(m)/(sec^2)`

As the acceleration is same for both the plane so a for second plane will be 2.67
`(m)/(sec^2)`

The another equation of motion is
`v^2=u^2+2as` using this equation for second plane
`40^2=0+2* 2.67* s`

s = 300 m