Q. Two particle A and B are moving in same direction on same straight line. A is ahead of B by 20m. A has constant speed 5 m/sec and B has initial speed 30 m/sec and retardation of 10 m/sec2. Then if  x(in m)  is total distance travelled by B as it meet A for second time.  Then value of x will be.

Dear student,
Distance moved by particle A is x1 = vtx1 = 5tDistance moved by particle Bis x2=ut-1/2 at2x2=30 t-1/2 (10) t2...1x1-x2=5t-30 t-1/2 (10) t220=-25 t+5t25t2-25t-20 =025±252-45×-2010=t25±625+40010=tt=5.7 sFrom 1 x2=30×5.7 -55.72x2=9x1=28.5
So x is 9 m.
Regards

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