An airplane originally at rest on a runway accelerates uniformly at 6.0 meters per second2 for 12 seconds. during this 12-second intervai, the airplane travels a distance of approximately

it would be 5,400n. i hope this !

solution

in this question, we have given,

acceleration=

acceleration=

acceleration=

we will first calculate displacement of jetliner when speed is constant  

from second equation of motion we know that,

now calculate displacement s' when it is decelerating  

so

therefore total displacement=s+s'

the final velocity of the skateboarder is 23.75 m/s.

we can use conservation of momentum to calculate the magnitude of the final velocity.

momentum before collision is equal to the momentum after collision

m1v1+m2v2=(m1+m2)v

here, v is final velocity of the skateboarder, assuming right side velocity as positive, and left side velocity as negative.

70*12+2000*(-25)=(70+2000)v

v=-23.75 m/s

here, negative sign indicates that both skyboarder and bus will move in left direction.