Question

Two particles A and B start moving simulteneously from point Pwith velocities 15 m/s and 20 m/s respectively.
They move with equal accelerations but opposite in direction. When A overtakes B at C then its velocity is
30 m/s. The velocity of B at C will be:
1) 5 m/s
2) 10 m/s
3) 15 m/s
4) 20 ms​

Answer 1

SOLUTION

GIVEN

• Initial velocity of A = 15m/s
• Initial velocity of B = 20m/s
• Final velocity of A = 30m/s

TO FIND

• Final velocity of B

EQUATION OF MOTIONS

$v=u+at$

$v² =u²-2aS$

$S = ut + ½at²$

ACCORDING TO QUESTION

$30= 15+at for P$

$V= 20+bt for Q$

$15= at$

$V- 20= bt$

$15/(V- 20)= a/b$

Now,

$900 -225= 2aS$

$675=2aS for P$

Again

$V^2- 400= 2bS$

$675/(V^2-400)= a/b$

So from above

$15/(V-20)= 675/(V^2-400)$

$(V-20)/15= (V^2-400)/675$

$V-20= V^2-400/45$

$V^2-400–45V+ 45*20= 0$

$V^2-45V+500=0$

$V^2-25V-20V+500= 0$

$V(V-25)—20(V- 25)= 0$

$V= 25 m/s.$Or

$V=20 m/s$

V must be 25m/s for B as B started itself with 20 m/s but with accln.

Answer 2

Answer:

Answer :

B

Solution :

<br> For Q,

Explanation:

Solution

Answer :

B

Solution :

<br> For Q,