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Question

A particle moving with a non-zero velocity & constant acceleration on straight line travels 15 m in first 3 s & 33 m in next 3 s in the same direction,find
(i) initial velocity of particle
(ii) acceleration of particle

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Solution

The particle travels a distance of 15 m in the first 3 s.
s1 = u1t+0.5at2
15 = 3u1+4.5a ...(1)
The particle travels a distance of 33 m in the second 3 s.
s2 = u2t+0.5at2
u2 = u1+at = u1 +3a
s2=33 = (u1+3a)3+4.5a = 9a+3u1+4.5a
33=9a +(3u1+4.5a) = 9a + 15
⇒33-15 = 9a
⇒18 = 9a
Hence, a = 2 m/s2
Putting the value of a in eq. (1), we get
15 = 3u1 +4.5×2
⇒3u1 = 15-9
⇒u1 = 6/3 = 2 m/s

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