Asked by Bridge
Two planes approach each other head-on. Each has a speed of 765 km/h, and they spot each other when they are initially 11.0 km apart. How much time do the pilots have to take evasive action?
Answers
Answered by
Henry
Since they are traveling at the same speed, each plane has the same amount of time and distance to take evasive action.
d = Vt,
t = d / V = (11/2) / 765 = 0.0072s.
d = Vt,
t = d / V = (11/2) / 765 = 0.0072s.
Answered by
Anonymous penguin
x=x0 + v0t +0.5at^2
Acceleration is zero in the x direction
My numbers are 800km/hr and 10.5 km
Divide the distance between the planes by two since traveling at same speed
0= 5.2 + 800t
Solve for t
t=23.4 seconds
Acceleration is zero in the x direction
My numbers are 800km/hr and 10.5 km
Divide the distance between the planes by two since traveling at same speed
0= 5.2 + 800t
Solve for t
t=23.4 seconds
Answered by
W
Two planes approach each other head-on. Each has a speed of 760 km/h
k
m
/
h
, and they spot each other when they are initially 12.3 km
k
m
apart.
k
m
/
h
, and they spot each other when they are initially 12.3 km
k
m
apart.
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