Question

A track is tangent to a circular track of radius r. Two particles A and B start simultaneously from a common point of track. Particle A moves with a uniform speed vo on the straight track whereas particle B moves on the circular track always collinear with centre of track and the point A. If the speed of B when it is equidistant from the centre and particle A is Vo then find the value of k.​

Answer 1

Answer:

6

Explanation:

Answer 2

Answer:

The velocity of particle B is $\frac{v_{0} }{4}$

Explanation:

A track is perpendicular to a track with a radius of r. A shared point of the track serves as the simultaneous launch point for particles A and B. While particle B goes on a circular track constantly collinear with the track's center and point A, particle A moves on a straight track at a constant speed of $v_{0}$. When particle B is equally far from the center and its speed is $V_{B}$.

So,

By given condition; we have

Rsecθ - R = R

Simplifying, we get,

2R = Rsecθ

secθ = 2

So, θ = 60°

Now, we have

tanθ = $\frac{x}{R}$

$v_{0}$ = R$sec^{2}$θ(dθ/dt)

$v_{0}$ = 4R × ω

$V_{B} = \frac{v_{0} }{4}$

Therefore, the velocity of particle B is $\frac{v_{0} }{4}$

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