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.

  • -21

avg. velocity is total distance/total time...

=200+220/2+5 =440/7 cm/s

ie. 4.4/7 m/s = 0.6285714....

  • -27
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