A747 jet, traveling at a velocity of 70 meters per second north, touches down on a runway. the jet slows to test at the rate of 2 meters per second^2

Distance, d = 1225 meters

Explanation:

It is given that,

Initial speed of the jet, u = 70 m/s

Acceleration of the jet,

It can be assumed to find the distance traveled on the runway as it is brought to rest. Let the distance is given by d. Also, the final speed, v = 0

d can be calculated using the third equation of kinematics. The formula is given by :

Substituting the values in above formula.

d = 1225 meters

So, the distance covered by the jet as it brought to rest is 1225 meters. Hence, this is the required solution.

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary 

its c i just checked it