Home
Class 12
PHYSICS
When two particeles A and B are at point...

When two particeles A and B are at point O, A is moving with a constant velocity 50 m/s , whle B is not moving. But B possesses a constant acceleration of 10 m/`s^(2)` . After how much time they will be at a distance of 125 m ?

Text Solution

Verified by Experts

For the particle A , ` U_(A) = 50 m//s , a _(A) = 0 m//s^(2)`
The particel B ` u_(B) = 0 , a_(B) =1-m//s^(2)`
So intial velocity of A w.r.t b
`u_(AB) = U_(A) -U_(B) = 50 ` m/s
and acceleration of A w.r.t B
` a_(AB) =a_(A) -a_(B) = -10 m//s^(2)`
The distance between A and B after time t is given by
` S_(AB) = U_(AB) t + 1/2 a_(AB)t^(2)`
` 125 = 50 t - 5t^(2)`
` or t^(2) - 10t + 25 =0`
` Rightarrow (t-5)^(2) =0`
` Rightarrow ` t= 5s.
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    AAKASH INSTITUTE|Exercise llustration 1 :|1 Videos

  • MOTION IN A PLANE

    AAKASH INSTITUTE|Exercise Try yourself|48 Videos

  • MOCK_TEST_17

    AAKASH INSTITUTE|Exercise Example|15 Videos

  • MOTION IN A STRAIGHT LINE

    AAKASH INSTITUTE|Exercise ASSIGNMENT (SECTION - D)|15 Videos

Similar Questions

Explore conceptually related problems

Two particles , one with constant velocity 50 m//s and the other with uniform acceleration 10 m//s^(-2) . Start moving simultaneously from the same place in the same direction. They will be at a distance of 125m other after.

Two cars A and B are at rest at the origin O. If A starts with a uniform velocity of 20 m// s and B starts in the same direction with a constant acceleration of 2 m//s^(2) , then the cars will meet after time

In the previous problem if the car is moving with a constant acceleration of 2 m//s^(2) , the ball will fall behind the person at a distance

A boy is moving with constant velocity 12 m//s . When he is 32 m behind a cyclist, the cyclist starts from rest and moves under constant acceleration 2 m//s^(2) . After how much time the boy meets the cyclist? Explain reason for two answers.

Two particles, one with constant velocity 50 m//s and the other with uniform acceleration 10m//s^(2) , start moving simultaneously from the same position in the same direction. They will be at a distance of 125m from each other after

Two particles, one with constant velocity 50 m/s and the other with uniform acceleration 10 m//s^(2) start moving simultaneously from the same place in the same direction. They will be at a distance of 125m from each other after:

Two cars A and B are at rest at same point initially. If A starts with uniform velocity of 40 m/sec and B starts in the same direction with constant acceleration of 4m//s^(2) , then B will catch A after :

Two particles A and B separated by 10 m at time t = 0 are moving uniformly. A is moving along line AB at a constant velocity of 4 m//s and B is moving perpendicular to the velocity of A at a constant velocity of 5 m//s . After what time the two particles will be nearest to each other?

Two particles A and B are initially 40 m apart, A is behind B . Particle A is moving with uniform velocity of 10 m s^(-1) towared B . Particle B starts moving away from A with constant acceleration of 2 m s^(-1) . The time which there is a minimum distance between the two is .

A cyclist starts from rest and moves with a constant acceleration of 1 m//s^(2) . A boy who is 48 m behind the cyclist starts moving with a constant velocity of 10 m//s . After how much time the boy meets the cyclist?