Two particles start their motion from M and N with velocity of 252km/h and 144km/h and acceleration of -4m/s2 and 8m/s2 respectively in the direction from M to N .Both the particles will meet each other twice at distance ______mand again after ______m. (MN=36m).
a. 148,228 b. 132,192 c. 252,144 d. 63,18

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Convert velocities to m/s:u1=252 km/h    =252 ×518 m/s    =70 m/su2=144 km/h    =144 ×518 m/s    =40 m/sLett=Time taken to meet each otherCondition to meet:Distance travelled by first particle=Distance travelled by 2nd particleu1t+12a1t2=distance of MN+u2t+12a2t270t+12×-4×t2=36+40t+12×8×t270t-2t2=36+40t+4t26t2-30t+36=0t2-5t+6=0By solving, t= 2 s and 3 sDistance of meeting:Substitute t= 2s in the equation 70t-2t2:s1=70×2-2×22    =132 mSubstitute t= 3s in the equation 70t-2t2:s2=70×3-2×32    =192 mAns:Option - b

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