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Two particles p and q are initially 40m apart p behind q. Particle p starts moving with a unifom velocity 10m/s towards q. Particle q starting from rest has an acceleration 2m/s^2 in the direction of velocity of p. Then the minimum distance between p and q will be.

shruti , 7 Years ago
Grade 10
anser 1 Answers
naveen

Last Activity: 7 Years ago

 
use v^2=u^2+2as
v is the final velocity,
u is the initial velocity,
a is the acceleration,
s is the minimum distance or displacement
find the time,
using v=u+at and t=5.0s AND s=ut+0.5ut^2 and u=5.0ms^-1therefore final answer is 22.75m using v^2=u^2+2as
 

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