A particle moves along the x axis according to the equation x= 2.00 + 3.00t 1.00t 2, where x is in meters and t is in seconds. At t= 3,00 s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration?
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Answer:
To find position we simply evaluate the given expression. To find velocity we differentiate it. To find acceleration we take a second derivative.
With the position given by x=2.00+3.00t−t
2
, we can use the rules for differentiation to write expressions for the velocity and acceleration as functions of time:
v
x
=
dt
dx
=
dt
d
(2+3t−t
2
)=3−2t and a
x
=
dt
dv
=
dt
d
(3−2t)=−2
Now we can evaluate x,v, and a at t=3.00s.
(a) x=(2.00+9.00–9.00)m=2.00m
(b) v=(3.00–6.00)m/s=–3.00m/s
(c) a=–2.00m/s
2
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