Question

5. A semicircle of radius R = 5m with diameter AD is sh
adius R = 5m with diameter AD is shown in figure.
Two particles 1 and 2 are at points A and B on shown diameter at
t = 0 and move along segments AC and BC with constant specu
V 2
-2mB
u, and u, respectively. Then the value of 41 for both particles to
U2
reach point C simultaneously will be:
(A) 5v2
(B) 202
(C) 2/2
tinle in moving in a circle of radius r with speed v as shown in
​

Answer 1

Answer:

√5/ 2

Explanation:

Let say O is the center of Circle

Then AO = Radius = 5 m

OB = AO - AB = 5 - 2 = 3 cm

OC = Radius = 5 m

in Δ OBC

BC² = OC² - OB²

=> BC² = 5² - 3²

=> BC² = 16

=> BC = 4

AC² = AB² + BC²

=> AC² = 2² + 4²

=> AC² = 4 + 16

=> AC² = 20

=> AC = 2√5

Distance = Speed * time

AC = 2√5 = u₁ * t

BC = 4 =  u₂ * t

=> u₁/u₂  = 2√5/4  =  √5/ 2

u₁/u₂  =  √5/ 2

Answer 2

Answer: √5/√4

Explanation:

We have,

Particle 1 and particle 2 moving across paths AC and BC respec+ively with velocity u_1 and u_2 respectively.

Since,

Radius = 5 m

=> OA = 5m

=> AB+BO = 5m

=> 2+BO = 5

=> BO = 3 m

In rt triangle OBC,

Radius = OC = 5m

BO = 3 m

By Pythagoras theorem,

$OC^2=BC^2+OB^2\\ \\=>BC^2 = 5^2-3^2=4^2\\ \\=>BC = 4m$

Also,

In triangle ABC,

Similarly,

$AC^2=BC^2+AB^2\\ \\=>AC^2=4^2+2^2\\ \\=> AC^2=20\\ \\=> AC=2\sqrt{5}m$

Since, both the particles are reaching the point C simultaneously.

=> distance travelled is directly proportional to speed.

$\frac{u_1}{u_2}=\frac{AC}{BC}\\ \\=\frac{2\sqrt{5}}{4}= \sqrt{ \frac{5}{4}}$