13) Fred is travelling 3.4 m/s and decelerates at a rate of 0.83 m/s until he comes to a stop How long does this take? How far does he travel during this time?

Tt will take 4.04 seconds

Explanation:

In this problem/exercise, we are going to apply the newtons first equation of motion to solve for the time taken

Given that

final velocity= 3.4m/s

initia velocity u= 0m/s

deceleration= 0.84 m/s^2

time t= ?

applying

v=u+at

Substituting our given data into the expression above we have

3.4=0+0.84t

3.4=0.84t

divide both sides by 0.84 we have

3.4/0.84= t

t= 4.04 seconds

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.