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Grade 11Modern Physics

A particle moving with uniform acceleration in a straight line covers a distance of 3m in the 8th second and the 5m in the 16th second of its motion .What is the displacement of the particle from the beginning of the 6th sec to the end of 15th sec?

Profile image of Rupesh patil
8 Years agoGrade 11
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1 Answer

Profile image of Vikas TU
8 Years ago
Dear Student,
From the Newton’s third law of eqn.
V2  - u2 / 2a = 3
=>6a= v2-u2

v=x+8a
u=x+7a

6a= 15a2+2ax
=>6=15a+2x

10a=(x+16a)2-(x+7a)2
=>10=31a+2x

10-6=(31a+2x)-(15a+2x)
=>4=16a
=>a=1/4 m/s2

Velocity at the start of 6th second = x+5a
=19/8 m/s

15/4+2x=6
=>x=9/8

Displacement = 19/8x11 + ½ x ¼ x 121
=41.25m.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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