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?
To determine the time the pilots have to take evasive action, we need to find the time it takes for the two planes to collide.
Let's break down the problem:
The two planes are traveling towards each other, so their combined relative speed is the sum of their individual speeds. In this case, both planes have a speed of 765 km/h, so the combined relative speed is 765 + 765 = 1530 km/h.
The pilots spot each other when they are initially 11.0 km apart. We need to find how much time it takes for the planes to cover this distance at their combined relative speed.
To find the time, we use the equation: time = distance / speed.
The distance is 11.0 km, and the speed is 1530 km/h.
Substituting these values into the equation, we have time = 11.0 km / 1530 km/h.
To find the time in hours, we divide the distance by the speed: time = 11.0 km / 1530 km/h = 0.0072 hours.
Finally, we can convert the time to minutes by multiplying it by 60: 0.0072 hours * 60 minutes/hour = 0.43 minutes.
Therefore, the pilots have about 0.43 minutes, or approximately 26 seconds, to take evasive action.
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.
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
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.