The two aircrafts are skidded for 14.9 s.
To find the answer, we need to know about the Newton's equation of motion.
<h3>What is the Newton's equation that relates velocity, distance, acceleration and time?</h3>
As per Newton's equation of motion
- V²-U²= 2aS
- V= U+at
- V= final velocity, U = initial velocity, S = distance, a= acceleration, t= time
<h3>What's the acceleration of the aircrafts that skidded from 35 m/s after the collision for 112.1 m before achieving 15 m/s?</h3>
- Here, U = 35 m/s, V = 15m/s, S= 112.1 m
- So, 15²- 35²= 2a×112.1= 224.2a
=> a= -1000/224.2= -4.5 m/s²
<h3>What's the time taken to stop if these aircrafts are de-accelerated by 4.5 m/s² from 35m/s?</h3>
- Here, U = 35 m/s, V=15 m/s a= -4.5 m/s²
- 35= 15+(-4.5)t
=> t= 20/4.5 = 4.4 s
Thus, we can conclude that the two aircrafts are skidded for 4.4 s.
Learn more about the Newton's equation of motion here:
brainly.com/question/25545050
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Answer:
As the speed is constant
Distance = Speed x Time
45min = (45/60)hrs = 0.75hrs
Distance = 75 x 0.75 = 56.25km
Explanation:
first you shouldnt give info as some people might log in and ruin your grades just post questions and you will get answers or just search the questions
If the object is moving at a constant speed, acceleration is 0. So:
The resultant force is 0. So the force pushing the object must be equal to the friction force acting, so
Friction =
90 N
