A force of 4.0N acts on a body of mass 2.0kg for 4.0s. Assuming the body to be initially at rest,Find the distance covered in 10s after the force starts acting?

Asked by  | 5th Sep, 2012, 12:24: PM

Expert Answer:

Acceleration acting on body for 4 sec. = 4/2 = 2m/ss
now distance covered in these 4 sec: s1 = 1/2(2)(4)2 = 16 m    (usin gthe second equation of motion)
now velocity after 4 sec = 2*4 = 8m/s   (using the first equation of  motion)
thus now the velocity will remain same = 8m/s
and the distance covered in remaining 6 sec = s2 = 6*8  = 48m
thus total distance = s1 + s2=  64m

Answered by  | 8th Sep, 2012, 06:23: PM

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