After landing a jet liner on a straight runway taxis to a st

After landing, a jet liner on a straight runway taxis to a stop at an average velocity of-35.0 km/h. If the plane takes 7.00 s to come to rest, what are the plane\'s initial velocity and acceleration?

Solution

avergae velocity = (initial velocity + final velocity ) / 2

as it stops, so final veocity vf= 0

35 = (vi + 0) /2

vi = 70 km/h .......Ans

vi = 70 x 1000 m / 3600 hr = 19.44 m/s

now applying vf = vi + a t

0 = 19.44 + 7a

a = 2.78 m/s^2 .....Ans