Asked by grace
given s(t) =1/(t+5) .. this is the position of the moving object.
Velocity?
Acceleration of the object 3 seconds after it starts moving?
Velocity?
Acceleration of the object 3 seconds after it starts moving?
Answers
Answered by
bobpursley
take the derivative to get velocity:
s'=v(t)=d/dt (t+5)^-1=-/(t+5)^2
a=v'(t)=-2/(t+5)^3 after three seconds = -2/8^3
s'=v(t)=d/dt (t+5)^-1=-/(t+5)^2
a=v'(t)=-2/(t+5)^3 after three seconds = -2/8^3
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