Question

A 1500 kg aircraft going 35 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and are going 15 m/s after the collision. If they skid for 112.1 m before stopping, how long (in seconds) did they skid? Hint: Are the aircraft moving at a constant velocity after the collision or do they experience an acceleration?

Answer 1

The two aircrafts are skidded for 14.9 s.

To find the answer, we need to know about the Newton's equation of motion.

What is the Newton's equation that relates velocity, distance, acceleration and time?

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

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?

• 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

What's the time taken to stop if these aircrafts are de-accelerated by 4.5 m/s² from 35m/s?

• 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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