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Question

The takeoff speed of a jetliner is 360 km/h. If the jetliner is to take off from a 2100 m long runway, what must be the (constant) acceleration along the runway?

Answer 1

(360km / 1 h) * (1000m / 1km) * (1 h / 3600 s) = 100 m/s = v_f

2 * a * d = v_f^2 - v_i^2

a = (v_f^2 - v_i^2) / (2 * d)

a = (100m/s ^ 2 - 0) / (2 * 2100m)

a = 100,000 / 4200

a = 2.381m/s^2

(2.381m/s^2) * (1 km / 1000m) * (3600^2s^2 / 1 h^2) = 30,857 km / h^2

simple check:

d = 0.5*a*t^2

t = sqrt(2*d/a)
t = sqrt(2*2100m/2.381m/s^2)
t = 42s

v = a*t
v = 2.381m/s^2 * 42s = 100m/s

Answer 2

Presuming the jetliner has a velocity of zero at time t=0 and that it instantly achieves the constant acceleration rate, you need to solve for the acceleration constant k which will result in reaching a velocity of 360 km/h = 360,000 m/h

Since there are 3600 seconds per hour, 360000/3600 = 100 m/s

by the time that x(t)=2100 m

x(t) = kt^2 = 2100 m

v(t) = 2kt = 100 m/s or k = 50/t

substitute 50/t for k into the first equation, giving

(50/t) * t^2 = 2100, t = 2100/50 = 210/5 = 42

substitute 42 for t into the first equation,

k * 42^2 = 2100, k = 2100/(42*42) = 2100/1764 = 525/441 =

(5*5*3*7)/(7*7*3*3) = 25/21 m/s^2

So k, constant acceleration rate = 25/21 = about 1.1904761904761904761904761904762 m/s^2

To check, substitute k = 25/21 and t = 42 into equation 1,

25/21 * 42^2 = 2100 m

and doing likewise into equation 2,

2 * 25/21 * 42 = 100 m/s

So with a constant acceleration of 25/21 m/s^2, the jetliner will reach takeoff speed of 100 m/s at exactly the time it reaches the end of the 2100 m runway, 42 seconds after beginning its takeoff roll.

I think that's right.

Answer 3

an acceleration of an additional 171.428 meters per hour every meter along the runway

Answer 4

2100 km =2.1km
360km=5.? per sec

Answer 5

who knows everyone knows math is for squares