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Grade: 11

                        

Two particles P and Q are initially 40m apart P behind Q. P starts moving with uniform velocity of 10m/s towards Q. Q starts from rest and has acceleration 2m/s2 in the dorection of P. Then minimum distance between P and Q will be ?

4 years ago

Answers : (1)

Vikas TU
12283 Points
							
From first law of eqn.
v = u + at
10= 0 + 2*t
t = 5s.
In this time
P would have moved 5*10 = 50 m  Q moves 5*5 = 25 m.
THERFEORE, minimum distance  it would cover = >40 + 25 – 50 = 15 m.
4 years ago
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