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Grade: 11

                        

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?

4 years ago

Answers : (2)

Vikas TU
12154 Points
							
For Partcle P,
let it covered the distance x then,
x + 40 = 10*t.............(1)
 
For Particle Q,
x  = 0 + 0.5*2*t^2................(2)
solve the both eqns. and get t and x values.
4 years ago
Mohit
11 Points
							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
						
3 years ago
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