# the distance between two particles is decreasing at the rate of 6m/s

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## The distance between two particle is decreasing at the rate of 6 m/sec. If these particles travel with same speeds and in the same direction, then the separation increase at the rate of 4 m/sec. The particle have speed as

Click here👆to get an answer to your question ✍️ The distance between two particle is decreasing at the rate of 6 m/sec. If these particles travel with same speeds and in the same direction, then the separation increase at the rate of 4 m/sec. The particle have speed as

Question

## The distance between two particle is decreasing at the rate of 6 m/sec. If these particles travel with same speeds and in the same direction, then the separation increase at the rate of 4 m/sec. The particle have speed as

**A**

## 5 m/sec ; 1 m/sec

**B**

## 4 m/sec ; 1 m/sec

**C**

## 4 m/sec ; 2 m/sec

**D**

## 5 m/sec ; 2 m/sec

Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is A)

Let the velocities be v and v’ respectively

Now, when the distance between two particles is decreasing we will have

v+v ′ =6m/s v=6−v ′ ....(I)

And when the distance is decreases

v−v ′ =4m/s v=4+v ′ .....(II)

Now, from equation (I) and (II)

4+v ′ =6−v ′ 2v ′ =2 v ′ =1m/s

Now, put the value of v’ in equation (II)

v=4+1 v=5m/s

Hence, the velocities 1 m/s and 5 m/s

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## The distance between two particles is decreasing at the rate of 6 m / sec. If these particles travel with same speeds and in the same direction, then the separation increase at the rate of 4 m / sec. The particles have speeds asA. 5 m/ cor ; 1 m/cB. 4 m/ ; 1 π/C. 4 m/ ; 2 mm/D. 5 m/ ; 2 m/s e c

The distance between two particles is decreasing at the rate of 6 m / sec. If these particles travel with same speeds and in the same direction, then the separation increase at the rate of 4 m / sec. The particles have speeds asA. 5 m/ cor ; 1 m/cB. 4 m/ ; 1 π/C. 4 m/ ; 2 mm/D. 5 m/ ; 2 m/s e c

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The distance between two particles is decreasing at the rate of 6 m / sec. If these particles travel with same speeds and in the same direction, then the separation increase at the rate of 4 m / sec. The particles have speeds asA. 5 m/ cor ; 1 m/cB. 4 m/ ; 1 π/C. 4 m/ ; 2 mm/D. 5 m/ ; 2 m/s e c

Question

The distance between two particles is decreasing at the rate of

6 m s e c

. If these particles travel with same speeds and in the same direction, then the separation increase at the rate of

4 m s e c

. The particles have speeds as

A B C D Open in App Solution

The correct option is **A**

When two particles moves towards each other then

v 1 + v 2 = 6 . . . . . ( i )

When these particles moves in the same direction then

v 1 − v 2 = 4 . . . . . ( i i ) By solving v 1 = 5 and v 2 = 1 m s Suggest Corrections 4

SIMILAR QUESTIONS

**Q.**The distance between two particles is decreasing at the rate of

6 m s e c

. If these particles travel with same speeds as in previous case and in the same direction, then the separation increase at the rate of

4 m s e c

. The particles have speeds as

**Q.**The distance between two particles is decreasing at the rate of

6 m s e c

. If these particles travel with same speeds as in previous case and in the same direction, then the separation increase at the rate of

4 m s e c

. The particles have speeds as

**Q.**The distance between two particle is decreasing at the rate of 6 m/sec. If these particles travel with same speeds and in the same direction, then the separation increase at the rate of 4 m/sec. The particle have speed as

**Q.**Two particles of equal mass have coordinates

( 2 m , 4 m , 6 m ) and ( 6 m , 2 m , 8 m )

. Of these one particle has a velocity

→ v 1 = ( 2 ^ i ) m/s

and another particle has velocity

→ v 2 = ( 2 ^ j ) m/s at time t = 0

. The coordinate of their centre of mass at time

t = 1 s will be

**Q.**Four particles of masses

1 kg , 2 kg , 3 kg and 4 kg

situated at the corners of a square start moving at speeds of

8 m/s , 6 m/s , 4 m/s and 2 m/s

respectively. Each particle maintains a direction towards the particle at the next corner symmetrically. Find the magnitude of velocity of the center of mass of the system at this instant.

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## The distance between two particles is decreasing at the rate of 6 m/sec. If these particles travel with same speeds and in the same direction, then the separation increase at the rate of 4 m/sec. The particles have speed as

When two particles moves towards each other then v1+v2=6 When these particles moves in the same direction then v1-v2=4 By solbing v1=5 and v2=1(m)/(s)

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The distance between two parti...

The distance between two particles is decreasing at the rate of 6 m/sec. If these particles travel with same speeds and in the same direction, then the separation increase at the rate of 4 m/sec. The particles have speed as

Updated On: 27-06-2022

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Text Solution Open Answer in App A 5 m/sec, 1 m/sec B 4 m/sec, 1 m/sec C 4 m/sec, 2 m/sec D 5 m/sec, 2 m/sec Answer

The correct Answer is A

Solution

When two particles moves towards each other then

v 1 + v 2 =6 v1+v2=6

When these particles moves in the same direction then

v 1 − v 2 =4 v1-v2=4 By solbing v 1 =5 v1=5 and v 2 =1 m s v2=1ms Answer

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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Guys, does anyone know the answer?