A body travels 200cm in the 1st 2 s and 220cm in the next 5 s.Calculate the velocity at the end of 7th second from start ? It's answer is 2.22 m/s
The displacement of the body in first 2 sec = 200cm
Let the initial velocity = u(say) ,
At t = 0, x(0) = 0, v (0) = u (say),
x (t) = 200cm, t = 2s
x(t') = (200 + 220)cm = 420cm
t' = (2 + 5)s = 7s
If a is the uniform acceleration of the particle,
then x(t) = x(0) + v(0)t + ½ at^2
200 = 0 + u*2 + 0.5*a *4
or 100 = u + a ........................................…(1)
Again, x(t') = x(0) + v(0)t' + ½ at^2
420 = 0 + u*7 + 0.5*a*49
60 = u + 3.5a ........................................… (2)
Subtracting (1) from (2),
-40 = 2.5a
or a = -16cm/s2
From equation (1),
u = 100 - (-16) = 116cm/s
Now, t'' = 7s, v(t'') = ?
v(t'') = v(0) + at'' v(t'') = (116 - 16*7) = 4cm/s.
A body travels 2m in first 2s.
Then, 2.2m in next 5s it shows that body is decelerating. So its velocity after 7s could not be 2.22m/s.