two particles P and Q are initially 40 m apart P behind Q. Particle P starts moving with a uniform velocity 10 m/s towards Q. Particle Q starting from rest has an acceleration 2 m/s² in the direction of velocity of P. Then the minimum distance between P and Q will be?
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Answered by
76
The minimum distance will be found when Q moves with 10m/s velocity which can be found byv=u+at10=0+2tt=5 sec.Distance traveled by P in 5 seconds is 5*10 i.e 50m. And distance traveled by Q in 5 seconds iss=0.5*2*(5)^2s=25m.So minimum distance between P and Q will be 40+25-50=15m
Answered by
111
the moment Q attains the velocity of 10 m/s, the time will correspnd to the minimum distance between the ii
=> v = u + at = 10 = 0 + 2(t)
=> t = 5 s
in 5 s P moves 5x10 = 50 m towards Q & Q moves a further 5x5 = 25 m
therefore min distance = 40 + 25 - 50 = 15 m
=> v = u + at = 10 = 0 + 2(t)
=> t = 5 s
in 5 s P moves 5x10 = 50 m towards Q & Q moves a further 5x5 = 25 m
therefore min distance = 40 + 25 - 50 = 15 m
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