Question

A body covers 100cm in first 2 seconds and 128 cm in next 4 seconds moving with constant acceleration. The velocity of the body at the end of the 8sec is

Answer 1

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,

$\purple{ \boxed{\sf{v=8\;cm/s}}}$

Answer 2

8 cm/s

Explanation:

Distance in first two seconds is

$s = ut + \frac{1}{2} a {t}^{2} \\ \\ 100 = 2u + \frac{1}{2} a {2}^{2} \\ \: \: 2u + 2t = 100 \\ u + t = 50$