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Old 17th April 2010, 10:02 AM
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Scoch Scoch is offline
Join Date: Sep 2009
Member of: Tayside Airgun Club
Location: Dundee, Scotland
Posts: 556

Originally Posted by Andy006 View Post
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
v=8.78m/s @ 18.27sec

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.


Andy the physicist

Yep thats what i thought too
Steve CochraneWalther LG300, Leupold MK4
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