Question

An aircraft has a landing velocity of 180 km/h. What length of runway is needed given that the deceleration of aircraft along the runway is 2.5 m/s^2?

a)100m
b)150m
c)320m
d)500m

Answer 1

Answer:

150m

Explanation:

because the runway of car is deceleration of aircraft

Answer 2

Answer:

(d) 500 m

Explanation:

First Convert km/h to m/s

To do multiply it with   $\frac{5}{18}$

∴ 180×$\frac{5}{18}$ = 50

by using formula $v^{2} = u^{2} + 2as$

⇒ $0^{2} = 50^{2} +2as$

⇒$0=2500+2(-2.5)s$             ∵ a is -ve since  it is decelerating  

⇒  $-2500=-5s$                         ∵ 2(-2.5)= 5                  

⇒$2500=5s$                                 ∵ cancel minus on both sides  

⇒ $\frac{2500}{5}$$= s$

⇒$500=s$  

⇒$s = 500$m                                      ∴ since the units are in meters  

∴ The length of runway required is 500 m

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