I think I understand Rob's theory.
A pellet starts from rest and accelerates at +1.87 m/s2 for 6.50 s. At the end of this time, the pellet continues for an additional 6.43 s with an acceleration of +0.656 m/s2. Following this, the pellet accelerates at -1.43 m/s2 for 5.34 s. (a) What is the velocity of the pellet at t = 18.27 s? (b) Find the total displacement of the pellet.
I got the answer to
a)
v=ta
v=6.5*1.87=12.155
v=6.43*.656=4.22
v=5.34*(-1.43)=-7.64
v=8.78m/s @ 18.27sec
b)
i cant seem to figure this out. i am trying displacement =vt, but this is getting me only 160, the answer is 198, but i dont know how to get 198
i just tried using this equation, and its not working out:
LaTeX Code: x = x_0 + v_0 t + (1/2) a t^2
for my first segment, my LaTeX Code: X_0 will be 0 right? but for my second segment the LaTeX Code: X_0 will be my answer to my first segment, correct?
I will do some more maths this afternoon, but I think Rob's on to something very special.
Quite easy really.
regards,
Andy the physicist