Question

An airplane lands on a runway with a velocity of 150 m/s. How far will it travel until it stops if its rate of

deceleration is constant -3m/s2?

Answer 1

3750 m

Explanation:

This question can be solved using law of equation of motion

`v^2 = u^2 + 2as`

where v is the final velocity

u is the initial velocity

a is the acceleration

and

s is the distance travelled

________________________________

given

final velocity = v = 0 since it is given that plane stops

u = 150 m/s

a = -3m/s2

using the above values in the given equation

`v^2 = u^2 + 2as\\0 = 150*2 + 2*-3s\\0 = 22500 - 6s\\=> 6s = 22500\\=> s = 22500/6 = 3750`

Thus, plane will travel 3750 m before it stops.