Question

an aeroplane lands on a runaway at a speed of 180km\hr and is brought to a stop uniformly in 30 secs . What distance does it cover before coming to rest​

Answer 1

The distance covered by the plane in its accelerated motion is 750 m

Explanation:

The motion of the plane is a uniformly accelerated motion, so we can use the following suvat equation:

$s=(\frac{u+v}{2})t$

where

v is the final velocity

u is the initial velocity

t is the time

s is the displacement

For the plane in this problem,

$u=180 km/h \cdot \frac{1000 m/km}{3600 s/h}=50 m/s$ is the initial velocity

v = 0 (the plane comes to a stop)

t = 30 s

Solving for s, we find the displacement of the plane, which corresponds to the distance covered:

$s=(\frac{50+0}{2})(30)=750 m$

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