a body covers 100cm in first 2 seconds and 128cm in the next four sec moving with a constant acceleration .find velocity of a body at the end of 8 sec?? pls ans it fast
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Explanation:
Answer:
Distance in first two seconds is
\implies \sf s_1 = ut_1 + \dfrac{1}{2} at^2
\sf{ 100 = 2u + \dfrac{1}{2}a(4) }\;---> (1)
Distance in (2+4) seconds from starting point is
\sf s_1+s_2 = u(t_1+t_2)+\dfrac{1}{2} a (t_1 + t_2)^2
\implies \sf{228 = 6u +\dfrac{1}{2}a(36) }--->(2)
From equation (1) and (2)
We get,
\blue{\boxed{\sf {a=-6\;cm/s^2}}}
Substitute \sf {a=-6}} in equation (1)
\sf \implies 100 = 2u - \dfrac{1}{2} (6)(4)
\implies \sf{2u = 112}
\green{\boxed{\sf u = 56\;cm/s}}
Using,
\pink {\boxed{\rm v = u + at}}
\implies \sf {56-(6*8)}
\implies \sf 56 - 48
Therefore,
\purple{ \boxed{\sf{v=8\;cm/s}}}
Answered by
1
-128 is the answer as I think
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