since s = vt + 1/2 at^2,
(10*6 + 0*6^2) + (10*10 + 4*10^2) = ____
Consider an object moving in the positive direction which passes the point x = 0 at time t = 0. Between t = 0 and t = 6 seconds, the object has a constant velocity of +10 m/s. At time t = 6 seconds, the object is given a constant acceleration of 8 m/s2 in the negative direction. What is the position of the object at t = 16 seconds?
4 answers
560 m?
also shouldnt the acceleration be negative?
Yes, the 4 should have a - sign
1 answer