1.

A body travels 200 čm in the first 2 s and220 cm in the next 5 s, Calculate the velocityat the end of the seventh second from thstart.(Ans. 2.22 m/

Answer»

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.


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