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Answer:

velocity at the end of 0.7sec from start would be 10cm/s

let initial velocity is u and acceleration is a.

case 1 : s₁ = 20 cm and t₁ = 0.2 sec

using formula, s₁ = ut₁ + 1/2 at₁²

⇒(20cm) = u × 0.2 + 1/2 × a × (0.2)²

⇒0.2 = 0.2u + 0.2 × 0.1 × a

⇒1 = u + 0.1a

⇒10 = 10u + a ............(1)

and velocity after 0.2 sec, v = u + at₁

= u + 0.2a

case 2 : s₂ = 22cm and t₂ = 0.4 sec

using formula, s₂ = vt₂ + 1/2 at₂²

⇒(22cm)² = (u + 0.2a) × (0.4) + 1/2 × a × (0.4)²

⇒0.22 = 0.4u + 0.08a + 0.08a

⇒0.22 = 0.4u + 0.16a

⇒11 = 20u + 8a ...........(2)

from equations (1) and (2) we get,

11 - 20 = 8a - 2a

⇒a = -9/6 = -3/2 = -1.5 m/s²

u = (10 - a)/10 = 11.5/10 = 1.15 m/s

velocity at the end of 0.7sec, v = u + a(0.7sec)

= 1.15 - 1.5 × 0.7

= 1.15 - 1.05

= 0.10

= 10cm/s

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