(a) From
v
(
t
) = 0 we find it is (momentarily) at rest at
t
= 0.
(b) We obtain
x
(0) = 4.0 m
(c) and (d) Requiring
x
(
t
) = 0 in the expression
x
(
t
) = 4.0 – 6.0
t
2
leads to
t
=
±
0.82 s for
the times when the particle can be found passing through the origin.
(e) We show both the askedfor graph (on the left) as well as the “shifted” graph which is
relevant to part (f). In both cases, the time axis is given by –3
≤
t
≤
3 (SI units
understood).
(f) We arrived at the graph on the right (shown above) by adding 20
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Physics

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