A body travels `200 cm` in the first `2 s` and `220 cm` in the next `4 s` with deceleration, The velocity of the body at the end of the seventhe second is .

Explanation

Let u be initial velocity and a be acceleration of the body. Since during the first two second, the distance traveled is s = 200 cm. Thus, we have
`s=ut +(1)/(2) at^(2)" or "200=2u +(1)/(2) xx a xx (2)^(2)" or "u+a=100" ….(i)"`
The distance covered in time t = 2 + 4 = 6s is s = 200 + 220 = 420 cm
Hence, `420 =6u +(1)/(2) xx axx (6)^(2)" or "u+3a =70" ....(ii)"`
Solving equations (i) and (ii), we get `u = 115 cm//s and a = – 15 cm//s^(2)`.
Therefore, the velocity at the end of the seventh second, i.e., at t = 7 s is `v = u + at = 115 –15 xx 7 = 10" cm s"^(–1)`