Calculate the acceleration of the particle at t = π/ 6 , if the position vector of the particle r = ( 5cosî + 2sin j + 6 t^ 2 sin 2 t k ) m
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Answer:
Position,
→
r
=94cos2t)
ˆ
I
+(4sin2t)
ˆ
j
+6t
ˆ
k
Velocity,
→
v
=
→
d
r
dt
=94(-sin2t).2}
ˆ
I
+[49cos2t).Hatj+6
ˆ
k
Acceleration,
→
a
→
d
v
9
dt)=[-89cos2t).2]
ˆ
I
+[8(-sin2t).2]
ˆ
j
=(-16cos2t0
ˆ
i
+(-16sin2t)
ˆ
j
Wher t=π/4,
→
a
=(-16cos2×π/4)
ˆ
i
+(-16sin2π/4)
ˆ
j
=(-16cosπ/2)
ˆ
i
+(-16sinπ/2)
ˆ
j
=(-16×0)
ˆ
i
+(-16×1)
ˆ
j
=-16
ˆ
j
ms-22.
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