Question

a ball travel a distance of 2m in 2 sec and 2.2 in 4 sec what will be the velocity of the body and the end of the 7sec from the start

Answer 1

s= ut + 1/2 a t^2 u is the initial velocity and a is the acceleration. 

The first relation you for 2 m 

2 = u 2 + 1/2 a 2 *2 

1= u + a 

For the second relation 

we get 

5 seconds after, i.e. 7 seconds after it covered the distance 2+2.2 

You get 

4.2 = u *7 + 1/2 a *7*7 

8.4 = 14 u + 49 a 

We got two relation 

u+a= 1 

14 u + 49 a = 8.4 

Sol;ve for a and u. You get 

u = 29/25 and a = -4/25 

The velocity V at the end of the 7th sec will be 

V= 29/25 - 4/25 *7 = 1/25 m/s