Question

70. Two particles P and Q simultaneously start movingfrom point A with velocities 15 ms and 20 ms -1respectively. The two particles move withaccelerations equal in magnitude but opposite indirection. When P overtakes Q at B then its velocityis 30 ms. The velocity of Q at point B will be(a) 30 ms-1 (b) 5 ms (c) 20 ms-(d) 15 ms?​

Answer 1

Answer:

5 m/s

Explanation:

→AP=15 m/s  

 →BP=30 m/s

AQ=20 m/s        

→BQ=v  

For P, v=u+at

⇒ at=v−u

⇒ at=30−15

⇒ at=15 m/s

For Q,v=u+(−at)v

=20−atv

=20−15

=5 m/s  

Answer 2

The Final velocity(V) of any particle in terms of its Initial velocity (U) , acceleration(a) and time taken (t) is given by:

                               $V = U + at$

For Particle P :

Given, Initial velocity (U) = 15 $ms^{-1}$

   V = 15 + at

Given final velocity (V) of particle P = 30 $ms^{-1}$

That is , 30 = 15 + at

Therefore , at = 15 $ms^{-1}$

For Particle Q:

Given, Initial velocity (U) = 20 $ms^{-1}$

at = 15 $ms^{-1}$   ( Since given acceleration is equal in magnitude)

V = U -at    ( Since acceleration opposite in direction of motion )

V = 20 - 15

V = 5 $ms^{-1}$

The velocity of Q at point B will be (b) 5 $ms^{-1}$. The correct option is (b) 5 $ms^{-1}$.

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