A jetliner lands at speed of 210 km/h and breaks at rate of 20 km/h/s . If the jatliner travels at a constant speed of 210 km/h for 1s after landing before applying the breakes . What is the total displacement of the jetliner between couchdown on the runway and coming to rest ?
To find the total displacement of the jetliner, we need to calculate the distance traveled while decelerating and the distance traveled while maintaining a constant speed for 1 second.
First, let's calculate the distance traveled during deceleration using the equation:
distance = (initial velocity)^2 / (2 * acceleration)
Given:
Initial velocity (v0) = 210 km/h
Acceleration (a) = -20 km/h/s (negative because it's deceleration)
Converting the units to meters per second:
Initial velocity (v0) = 210 km/h = (210 * 1000) m/3600 s = 58.33 m/s
Acceleration (a) = -20 km/h/s = (-20 * 1000) m/3600 s = -5.56 m/s^2
Using the formula for distance,
distance = (v0^2) / (2 * a)
distance = (58.33^2) / (2 * -5.56)
distance = 679.16 meters (rounded to two decimal places)
So, during the deceleration phase, the jetliner travels 679.16 meters.
Next, let's calculate the distance traveled while maintaining a constant speed of 210 km/h for 1 second.
Speed (v) = 210 km/h
Time (t) = 1 second
Converting the units to meters per second:
Speed (v) = 210 km/h = (210 * 1000) m/3600 s = 58.33 m/s
Time (t) = 1 second
Using the formula for distance,
distance = v * t
distance = 58.33 * 1
distance = 58.33 meters
So, while maintaining a constant speed for 1 second, the jetliner travels an additional 58.33 meters.
Finally, we can find the total displacement by adding the distances traveled during the deceleration phase and the constant speed phase:
Total displacement = distance during deceleration + distance during constant speed phase
Total displacement = 679.16 meters + 58.33 meters
Total displacement = 737.49 meters (rounded to two decimal places)
Therefore, the total displacement of the jetliner between touchdown on the runway and coming to rest is 737.49 meters.