Question

9. A jet plane lands with a velocity of 200m/s and can decelerate at a maximum rate of 10.0m/s2 as it comes to rest. (a) From the instant it touches the runway, how much time is needed for the plane to come to rest? (b) What is the minimum length of runway needed for this plane to land? (c) The same plane needs to reach a speed of 82.1m/s in order to take off from this exact runway (same length as part b). What minimum acceleration is required for the plane to take off? (d) How much time has elapsed before this plane takes off?

Answer 1

20s ; 2000m(2km) ; 1.69m/s^2 ; 48.7s

Step-by-step explanation:

Velocity (U) = 200m/s

deceleration( negative acceleration) (a) = 10m/s^2

Final Velocity (V) (rest) = 0

A) time needed to come to rest, V =0

Using

V = u + at

Where t = time

0 = 200 - 10t

10t = 200

t = 200/10

t = 20s

B) minimum length of runway required :

Using, v^2 = u^2 + 2aS

Where S = distance(Length) in metres

0 = 200^2 + 2×(-10)×S

0 = 40000 - 20S

20S = 40000

S = 40000/20

S = 2000m = 2km

If plane requires speed of 82.1m/s to take off on same runway:

Minimum take-off acceleration :

v^2 = u^2 + 2aS

82.1^2 = 0 + 2×a×2000

4000a = 6740.41

a = 6740.41 / 4000

a = 1.685m/s^2

Time elapsed before taking off

S = ut + 0.5at^2

2000 = 0 + 0.5(1.685)t^2

2000 = 0.8425t^2

t^2 = 2000/0.8425

t^2 = 2373.8872

t = 48.7s