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

Answer:

300 m

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 $S=ut+\frac{1}{2}at^2$ where s is distance a is acceleration and u is initial velocity

Using this equation for first plane $1200=0\times 30+\frac{1}{2}a30^2$

$a=2.67\frac{m}{sec^2}$

As the acceleration is same for both the plane so a for second plane will be 2.67 $\frac{m}{sec^2}$

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

s = 300 m