physics: acceleration problem?

Question

A Cessna aircraft has a lift-off speed of 120 km/h. (a) What minimum constant acceleration does this require if the aircraft is to be airborne after a takeoff run of 250m?
m/s2
(b) How long does it take the aircraft to become airborne?
s

the webiste says that 2.22 m/s62 and 15 s is wrong!

m/s^2 lol sry

Answer 1

The answer are these: the minimum acceleration is 2.222... m/s^2, and the time taken is 15 seconds.

Here's how these results were found:

120 km/h = 120*1000/(3600) m/s = 33.33... m/s

(a) v^2 = 2ad implies that a = v^2/(2d) = (33.33...)^2 / (500) m^2/s^2

= 2.222... m^2/s^2.

(b) For constant acceleration with a final speed v,

d = 1/2 v t, so that t = 2d / v = 500 / 33.333... s = 15 s.

Live long and prosper.

Re. your "additional details": the "webiste" is rwong!

Answer 2

If the boy drops the ball to free fall, the following should apply Initial Velocity is 0 m/s Final Velocity is 14.7 m/s Acceleration is (gravity =9.8 m/s2) distance = 1/2 * gt^2 = 1/2 * 9.8 * 4 = 19.6 m If the boy power throws the ball with initial speed of 14.7 m/s, then the following should apply d = Vi *t + 1/2 gt^2 d= 14.7*2 + 1/2*9.8*4 = 49 m

Answer 3

For part a use the relationship:

x = (v^2-v0^2)/(2a) and solve for a.

a = (v^2-v0^2)/(2x) x = 250 m, v0 = 0, v = 120km/hr

v = 120 km/hr = 33.333m/s

a = 2.22 m/s^2

b. Use v = at ---> t = v/a = 33.33 m/s/2.22m/s^2 = 15 sec

Answer 4

Initial Velocity squared- final velocity squarted divided by two times the distance

Vi2-V02/2X = (33.33)2/480 = 2.31m/s2

Answer 5

120km/h(1000m/1km)(1hr/3600s) to convert it to m/s.

Then vf^2=vi^2+2ad.
Where vi^2 is really 0.
So vf^2=2ad.
Therefore vf^2/(2(d))=a.

Then vf=vi+at.
And vi^2 is again 0.
So vf^2/a=t.

Where vf is velocity final, vi is velocity initial, a is acceleration, and t is time.