SOLUTION: Starting from rest, a particle moving in a straight line has an acceleration,π = (ππ β π)π/ππ, where t is in seconds. What is the particleβs velocity when
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Question 1182534: Starting from rest, a particle moving in a straight line has an acceleration,π = (ππ β π)π/ππ, where t is in seconds. What is the particleβs velocity when t= 6s, and what is its position when t = 11 s? At what time will it reach a velocity of 2 m/s?
Found 2 solutions by Solver92311, ikleyn:
Answer by Solver92311(821) (Show Source): You can put this solution on YOUR website!
\ =\ \int\,a(t)\,dt\ +\ v_o)
\ =\ \int\,\(v(t)\,+\,v_o\)\,dt\ +\ s_o)
\ =\ \int\,(3t\,-\,5)\,dt\ +\ 0\ =\ \frac{3t^2}{2}\ -\ 5t)
\ =\ \frac{3(6)^2}{2}\ -\ 5(6))
\ =\ 24)
m/s
\ =\ \int\,\frac{3t^2}{2}\ -\ 5t\ +\ 0\ =\ \frac{t^3}{2}\ -\ \frac{5t^2}{2})
\ =\ 363)
m
\ =\ 2)
Solve for 
CORRECTED
John
My calculator said it, I believe it, that settles it
From
I > Γ
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
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