4.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?

200 Views | Posted 5 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

4.8 The radius of the ground = 200 m, the distance PQ = diameter of the ground = 400 m

The displacement all 3 girls = 400 m

This magnitude is equal to the path skated by girl B

0 Upvotes 0 Downvotes

Similar Questions for you

A ball is projected from ground at angle q with the horizontal. After t = 1 s, it is moving at 45° with the horizontal and after t = 2 second, it is moving horizontally. What is speed of projection of ball? [g = 10 m s^–2]

alok kumar singh

Please find the solution below:

An electron in a research apparatus follows a circle path. On the electron’s first circuit of the apparatus, its speed is v₀ and the radius of its circular path is R. Each time it makes on circuit, it passes a short region where it receives a “kick” and gains an additional speed of $\frac{v_{0}}{100} .$ The electron follows a circular path such that the magnitude of its acceleration is always the same. What is the radius of the circular path after the electron has received 10 kicks?

Vishal Baghel

$a = \frac{v_{0}^{2}}{R}$ $\frac{_{}^{}}{}$

after 10 kicks, $v = v_{0} + \frac{10 v_{0}}{100} = 1.1 v_{0}$ $_{} \frac{_{}}{}_{}$

$\Rightarrow R_{1} = 1.21 R$

Two particles start moving from the same point in two different directions. The first particle moves along the positive x-axis with a constant acceleration of 3 m/s². The second particle moves in a straight line making an angle of 60° with the positive x-axis with a constant speed of 24 m/s. Find the time after which the relative velocity of the particles is minimum.

Vishal Baghel

v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec

A particle is moving in a circle of radius 1m with angular velocity (ω) given as a function of angular displacement ( θ ) by the relation ω = θ² + 2θ. The total acceleration of the particle when θ = 1rad is:

Vishal Baghel

ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²

A flying disc of radius r is moving with constant speed v? while rotating anticlockwise with angular speed v?/r along a curve PQ as shown below. Radius of curvature of curve at the instant is 4r. The magnitude of acceleration of point A at the instant is (3v₀²)/(nr). Find n.

Vishal Baghel

a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r