Asked by Ram kumar
A particle is moving along a straight line and its position is given by the relation x=( t3-6t2-15t+40)mm. Find:-
(a). The time at which velocity is zero.
(b). Position & displacement of the particle at that point.
(c). Acceleration for the particle at that line.
(a). The time at which velocity is zero.
(b). Position & displacement of the particle at that point.
(c). Acceleration for the particle at that line.
Answers
Answered by
MathMate
Given a position function x(t), then
x'(t)=dx(t)/dt=velocity, and
x"(t)=d²x(t)/dt²=acceleration.
where x' and x" are the first and second derivatives with respect to time.
(a) solve for x'(t0)=0
(b) evaluate position = x(t0), and displacement = x(t0)-x(0).
(c) evaluate x"(t0)
x'(t)=dx(t)/dt=velocity, and
x"(t)=d²x(t)/dt²=acceleration.
where x' and x" are the first and second derivatives with respect to time.
(a) solve for x'(t0)=0
(b) evaluate position = x(t0), and displacement = x(t0)-x(0).
(c) evaluate x"(t0)
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