Question

A body travels 200 cm in the frst two seconds and 220cm in the next 4 sec with same accerlartion the velocity of the body at the end of the 7th second is

Answer 1

let initial velocity be u and acceleration be a

s = ut+ 1/2 (at^2)

EQUATION 1:

200 cm = 2 m

s = 2m

t= 2 sec

2 = u x 2 + 1/2 (a x 2^2)

2 = 2u + 2a

u + a =1 ..........................equation 1

EQUATION 2:

220 cm = 2.2 m

s = 2 + 2.2 = 4.2 m

t= 2 + 4 = 6 sec

4.2 = u x 6 + 1/2 (a x 6^2)

4.2 = 6u + 18 a

0.7 = u + 3a..........................equation 2

solving equations 1 and 2,we get

a= (-0.15)m/s^2   and    u = 1.15 m/s

now for the velocity at the end of 7 sec

v = u + at

v = 1.15 + (-0.15) x 7

v= 1.15 - 1.05

v = 0.1 m/s   (or)   10 cm/sec

hope this helps