Asked by Anon
A particle of mass 45 kg moves in a straight line such that the force (in Newtons) acting on it at time t (in seconds) is given by
225t^4-90 t^2-225.
If at time t = 0 its velocity, v , is given by v ( 0 ) = 2, and its position x (in m) is given by x ( 0 ) = 5, what is the position of the particle at time t?
225t^4-90 t^2-225.
If at time t = 0 its velocity, v , is given by v ( 0 ) = 2, and its position x (in m) is given by x ( 0 ) = 5, what is the position of the particle at time t?
Answers
Answered by
bobpursley
F=Ma
a=F/m=225/45 t^4-90/45 t^2-225/45
a=5t^4-2t^2-5
v=int a(t)dt
=t^5-2/3 t^3-5t+C
given at t=O, v=2
C=2
if v=t^5-2/3 t^3-5t+2 then
x=INT v(t)dt
x=1/6 t^6+1/6 t^4 +2.5t^2+2t+C
given x(o)=6 implies then C=5
a=F/m=225/45 t^4-90/45 t^2-225/45
a=5t^4-2t^2-5
v=int a(t)dt
=t^5-2/3 t^3-5t+C
given at t=O, v=2
C=2
if v=t^5-2/3 t^3-5t+2 then
x=INT v(t)dt
x=1/6 t^6+1/6 t^4 +2.5t^2+2t+C
given x(o)=6 implies then C=5
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