Question

A body travels 200 cm in the first 2 s and
220 cm in the next 5 s. Calculate the
velocity at the end of the seventh second from
the start.​

Answer 1

Answer:

Here is the answerThe 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 + � at^2 200 = 0 + u*2 + 0.5*a *4 or 100 = u + a ........................................…(1) Again, x(t') = x(0) + v(0)t' + � 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.