Question

integrate it plzzz... ​

Answer 1

$\Huge{\sf Kinematic \ Equations :}$

These equations are only to be applied when an object is executing Uniform Motion

We have to derive s = ut +½at²

W.K.T,

We would need displacement in our equation. So,

$\sf \: v = \dfrac{ds}{dt} \\ \\ \longrightarrow \: \sf \: ds = vdt$

But v = u + at

$\longrightarrow \: \sf \: ds = (u + at)dt$

Applying Integral on both sides of the equation,

$\longrightarrow \displaystyle \sf \: \int^{s}_{0}s \: = \int^{t}_{0} u.dt + \int^{t}_{0}at.dt \\ \\ \longrightarrow \: \sf \bigg[s \bigg]^{s}_{0} \: = u\bigg[t \bigg]^{t}_{0} + a\bigg[ \dfrac{ {t}^{2} }{2} \bigg]^{t}_{0} \\ \\ \longrightarrow \: \sf \: s - 0 = u(t - 0) + \dfrac{1}{2} a( {t}^{2} - 0) \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: s = ut + \frac{1}{2} at {}^{2} }}$

$\rule{300}{2}$

We have to derive v² - u² = 2as

W.K.T,

$\sf \: a = \dfrac{dv}{dt} \\ \\ \longrightarrow \: \sf \: a = \dfrac{dv}{dt} \times \dfrac{ds}{ds} \\ \\ \longrightarrow \: \sf \: a = v \dfrac{dv}{ds} \\ \\ \longrightarrow \: \sf \: a.ds = v.dv$

Integrating on both sides,

$\longrightarrow \: \displaystyle \sf \int^v_u \: v.dv = a\int^s_0ds \\ \\ \longrightarrow \: \sf \: \bigg[ \dfrac{ {v}^{2} }{2} \bigg]^v_u = a \bigg[s \bigg]^s_0 \\ \\ \longrightarrow \sf {v}^{2} - {u}^{2} = 2a(s - 0) \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: {v}^{2} - {u}^{2} = 2as}}$

$\rule{300}{2}$

$\rule{300}{2}$

Answer 2

Step-by-step explanation:

helpful

plz brainliest