An object moves along a straight line in such a way that it is slowing down with an acceleration of -8 meters per second per second. The object has an initial velocity of 14 meters per second. Find the distance the object travels during the first 3 seconds.

Question

anonymous · Answer

s=position    v= velocity   a=acceleration  t=time
$$s=s \iota+v \iota \Delta t+1/2a(\Delta t)^{2}$$

anonymous · Answer

s=0+(14m/s)(3)+1/2(-8)(3^2)
s=42-36
s=6meters

anonymous · Answer

But where did you get the equation you plugged the values in to? Did you derive it or just look it up? And if so how?