Question

A particle is at point A(1 m, 2 m, 2 m). Then the angle between its position vector and positive X-axis is​

Answer 1

Explanation:

given position vector of particle at time t

r

={(2t)i+(2t

2

)j}(m)

x=2t

⇒V

x

=

dt

dx

=2

y=2t

2

⇒V

y

=

dt

dy

=4t

tanθ=

V

x

V

y

⇒

2

4t

=2t

By differentiating with respect to time

dt

d

(tanθ)=

dt

d

(2t)

⇒(sec

2

θ)

dt

dθ

=2

⇒(1+tan

2

θ)

dt

dθ

=2

as we know from above tanθ=2t

⇒(1+4t

2

)

dt

dθ

=2

⇒

dt

dθ

=

1+4t

2

2

At t=2sec

⇒

dt

dθ

=

1+4×4

2

=

17

2

rad/s

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Answer 2

Answer:

given position vector of particle at time t

r

={(2t)i+(2t

2

)j}(m)

x=2t

⇒V

x

=

dt

dx

=2

y=2t

2

⇒V

y

=

dt

dy

=4t

tanθ=

V

x

V

y

⇒

2

4t

=2t