Question

[" (i) Derive the following relationships: "],[[" (i) "s=ut+(1)/(2)at^(2)," (ii) "v=u+at," (iii) "v^(2)=u^(2)+2as]]​

Answer 1

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Here's the velocity - time graph of a particle. Let the particle have a velocity u ms^(-1) at the time 0 second and v ms^(-1) for a time t second, i.e., v(0) = u and v(t) = v.

Here the slope of the graph gives the acceleration of the particle, i.e.,

$a=\dfrac{v-u}{t-0}\\\\\\a=\dfrac{v-u}{t}\\\\\\v-u=at\\\\\\\large\boxed{\mathbf{v=u+at}}$

Let $v-u=\Delta v$ and $t-0=\Delta t,$ i.e., change in velocity and time respectively.

Hence we have $a=\dfrac{\Delta v}{\Delta t}.$

But acceleration exists for the particle even as the change in time tends to be empty, i.e., ∆t → 0. So we have,

$\displaystyle a=\lim_{\Delta t\to 0}\dfrac{\Delta v}{\Delta t}\\\\\\a=\lim_{\Delta t\to 0}\dfrac{v-u}{\Delta t}\\\\\\a=\lim_{\Delta t\to 0}\dfrac{v(t)-v(0)}{\Delta t}$

But $t-0=\Delta t\quad\implies\quad t=0+\Delta t.$ So,

$\displaystyle a=\lim_{\Delta t\to 0}\dfrac{v(0+\Delta t)-v(0)}{\Delta t}$

Now we have the standard equation of derivative, so,

$a=\dfrac{du}{dt}$

Well, the area between the graph and the time axis, i.e., x axis, gives the displacement of the body. Since our graph is a straight line, we can directly give the area of the trapezium below the graph. If it's a curved one, we may integrate displacements occurring for a small time 'dt' seconds.

Well, integration is better.

From our graph, the displacement of the particle travelled in 'dt' seconds is,

$ds=u\ dt\quad\iff\quad u=\dfrac{ds}{dt}$

So the total displacement travelled by the particle between the time 0 seconds and t seconds will be,

$\displaystyle s=\int\limits_0^tu\ dt$

But $a=\dfrac{du}{dt}\quad\implies\quad dt=\dfrac{du}{a}.$

So 'dt' is replaced in terms of 'du'. Note the change in limits too.

$\displaystyle s=\int\limits_u^vu\dfrac{du}{a}\\\\\\s=\dfrac{1}{a}\int\limits_u^vu\ du\\\\\\s=\dfrac{1}{a}\left[\dfrac{u^2}{2}\right]_u^v\\\\\\s=\dfrac{v^2-u^2}{2a}\quad\longrightarrow\quad (1)\\\\\\v^2-u^2=2as\\\\\\\large\boxed{\mathbf{v^2=u^2+2as}}$

Consider (1) again.

$s=\dfrac{v^2-u^2}{2a}$

Since $v=u+at,$

$s=\dfrac{(u+at)^2-u^2}{2a}\\\\\\s=\dfrac{u^2+2uat+a^2t^2-u^2}{2a}\\\\\\s=\dfrac{t(2u+at)}{2}\\\\\\s=\dfrac{2ut}{2}+\dfrac{at^2}{2}\\\\\\\large\boxed{\mathbf{s=ut+\dfrac{1}{2}at^2}}$