    Question

# A particle starts from rest and has an acceleration of 2 m/s2 for 10 sec. After that, it travels for 30 sec with constant speed and then undergoes a retardation of 4 m/s2 and comes back to rest. The total distance covered by the particle is

A
650 m
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B
750 m
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C
700 m
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D
800 m
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Solution

## The correct option is B 750 mGiven Initially u=0 m/s;a=2 m/s2;t=10 s Velocity after 10 s Using 1st equation of motion v=u+at v=0+2×10=20 m/s Distance covered in this time Using 2nd equation of motion s=ut+12at2 s1=0+12×2×102=100 m After 10 s velocity is 20 m/s and it moves with this velocity for another 30 s Distance covered during this 30 s s2=20×30=600 m At last it retards from velocity 20 m/s to 0 m/s with deceleration of 4 m/s2 Distance covered during retardation Using 3rd equation of motion v2=u2+2as 02=202−2×4×s3 s3=50 m Total distance travelled =s1+s2+s3=100+600+50=750 m  Suggest Corrections  1      Similar questions
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