At time t1 = 14 s, a car is located at 99, 80, 27 m and has velocity 4, 0, - 3 m/s. At time t2 = 18 s, what is the position of the car? (The velocity is constant in magnitude and direction during this time interval.)

115, 80, 15m

Explanation

t1 = 14s

t2 = 18s

change in time = 4s (18-14)

r (final) = r (initial) + (average velocity) x (change in time)

multiply the average velocity with the change in time

= (4, 0, - 3) x 4 = 16, 0, - 12

now we'll add this value to the initial position of the car

(99, 80, 27) m + (16, 0, - 12) m = (115, 80, 15) m