Please tell how to solve numerical 15

Dear Student,

Total distance covered(s)=Distance during acceleration(s1)+ distance during uniform motion(s2)+distance during retardation(s3)for acceleration,v=u+atvmax=0+2×10=20 m/ss=ut+12at2s1=0×10+12×2×102=100 mFor uniform motion,s=vts2=20×200=4000 mDuring retardation,v=u+at0=20+a×50a=-0.4 m/s2retardation=0.4 m/s2s3=ut+12at2s3=20×50+12-0.4×502=500 mTotal distance, s=100+4000+500=4600 mtotal time, t=10+200+50=260 sec.Average velocity, vavg=total displacementtotal timevavg=4600260=17.69 m/s

Regards

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