The Eigen vectors of matrix :
coso - sin 07
4-sino cose
are :
Answers
Answer:
From the Question,
Mass of the particle,m = 3 Kg
Displacement of the particle is defined as:
\begin{lgathered}\large{\sf{s = \frac{1}{3} t {}^{2}}} \\\end{lgathered}
s=
3
1
t
2
Differentiating s w.r.t to t,we get velocity of the particle:
\begin{lgathered}\large{\sf{v = \frac{ds}{dt} }} \\ \\ \rightarrow \ \sf{v = \frac{d( \frac{1}{3}t {}^{2} ) }{dt} } \\ \\ \rightarrow \: \huge{ \rm{v \: = \frac{2}{3}t \: ms {}^{ - 1} }}\end{lgathered}
v=
dt
ds
→ v=
dt
d(
3
1
t
2
)
→v=
3
2
tms
−1
Differentiating v w.r.t to t,we get:
\begin{lgathered}\large{ \sf{a = \frac{dv}{dt}}} \\ \\ \rightarrow \ \sf{a = \frac{d(\frac{2}{3}t)}{dt}} \\ \\\huge{\rightarrow \: \boxed{\rm{a = \: {\frac{2}{3}} ms^{ - 2} }}}\end{lgathered}
a=
dt
dv
→ a=
dt
d(
3
2
t)
→
a=
3
2
ms
−2
We Know that,
F = ma
→ F = ⅔(3)
→F = 2 N
When t = 2, displacement would be:
\begin{lgathered}\sf{s = \frac{1}{2} (2) {}^{2}} \\ \\ \huge{\implies \: \sf{s = \frac{4}{3}m}}\end{lgathered}
s=
2
1
(2)
2
⟹s=
3
4
m
\therefore∴
\Huge{\boxed{\boxed{\sf{W = F.s}}}}
W=F.s
Putting the values,we get:
\begin{lgathered}\sf{W = 2 \times \frac{4}{3} } \\ \\ \implies \ \huge{\sf{W = \frac{8}{3} J}}\end{lgathered}
W=2×
3
4
⟹ W=
3
8
J
Thus,the work done by the force is 8/3 Joules
The correct option is____________(c)