What is the smallest distance, in meters, needed for an airplane touching the runway with a velocity of 360 km/h and an acceleration of -10 m/s^2 to come to rest?

Explanation

a) In this problem the initial velocity u = 360 km/h,
the final velocity v = 0 (rest) and
the acceleration a = -10 m/s^2.
The distance x can be calculated using the formula
`v^2 = u^2 + 2 a x`
Convert 360 km/h into m/s:360 (km)/h = (360 000 m)/(3600 s) = 100 m/s