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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?

Rupesh patil , 8 Years ago
Grade 11
anser 1 Answers
Vikas TU

Last Activity: 8 Years ago

Dear Student,
From the Newton’s third law of eqn.
V- 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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