A jet plane lands with a velocity of 200m/s and can decelerate at a maximum rate

Question

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? 9.

Explanation / Answer

Part C.

For this part

Initial Speed = 0 m/sec

final required speed for take-off = 82.1 m/sec

length of runway = 2000 m

Using 3rd kinematic equation

V^2 = U^2 + 2*a*d

a = (V^2 - U^2)/(2*d)

a = (82.1^2 - 0^2)/(2*2000)

a = 1.68 m/sec^2

Your mistake was that plane will be initially at rest, Since plane has come to rest. So to start again V0 = 0

Part D.

time taken by plane for take off will be

Using 1st kinematic equation

V = U + a*t

t = (V - U)/a

t = (82.1 - 0)/1.68

t = 48.87 sec

Again same mistake as in part C.

Please Upvote.

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