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A cessna 150 aircraft has a lift-off speed of approximately 125 km/h.

a) What minimum constant acceleration does this require if the aircraft is to be airborne after a take-off run of 206m? Answer in units of m/s^2.

b) What is the corresponding take-off time? Answer in units of s.

c) If the aircraft continues to accelerate at this rate, what speed will it reach 32.8 s after it begins to roll? Answer in units of m/s.

2007-09-17 11:13:17 · 1 answers · asked by grouchy187 2 in Science & Mathematics Physics

1 answers

The equations for this are
v(t)=a*t
and
x(t)=.5*a*t^2

From the statement
125/3.6 m/s=a*t
and
206=.5*a*t^2
plug a*t into the second equation from the first
125/3.6=.5*t
so t=250/3.6
or 69.4 seconds
so
125/3.6 =a*250/3.6
a=.5 m/s^2

c) v(32.8)=.5*32.8
or 16.4 m/s

j

2007-09-17 11:46:51 · answer #1 · answered by odu83 7 · 0 0

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