Question

An object is moving with uniform acceleration. Derive
an expression of the distance covered by the object in
nth second​

Answer 1

Let the uniform acceleration be $\displaystyle\sf {a,}$ which is the first derivative of velocity, i.e.,

$\displaystyle\longrightarrow\sf {\dfrac {dv}{dt}=a}$

$\displaystyle\longrightarrow\sf {dv=a\ dt}$

Let the initial velocity be $\displaystyle\sf {v_0}$ and the velocity attained at $\displaystyle\sf {n^{th}}$ second be $\displaystyle\sf {v.}$ Then,

$\displaystyle\longrightarrow\sf {\int\limits_{v_0}^vdv=\int\limits_0^na\ dt}$

$\displaystyle\longrightarrow\sf {\big[v\big]_{v_0}^v=a\big[t\big]_0^n}$

$\displaystyle\longrightarrow\sf {v-v_0=a\big[t\big]_0^n}$

$\displaystyle\longrightarrow\sf {v=v_0+a\big[t\big]_0^n}$

But we know that the velocity is the first derivative of the displacement, i.e.,

$\displaystyle\longrightarrow\sf {\dfrac {dx}{dt}=v_0+a\big[t\big]_0^n}$

$\displaystyle\longrightarrow\sf {dx=\left(v_0+a\big[t\big]_0^n\right)\ dt}$

$\displaystyle\longrightarrow\sf {\int\limits_0^sdx=\int\limits_0^n\left(v_0+a\big[t\big]_0^n\right)\ dt}$

$\displaystyle\longrightarrow\sf {\big[x\big]_0^s=\int\limits_0^nv_0\ dt+a\int\limits_0^nt\ dt}$

$\displaystyle\longrightarrow\sf {s-0=v_0\big[t\big]_0^n+a\left [\dfrac {t^2}{2}\right]_0^n}$

$\displaystyle\longrightarrow\sf {s-0=v_0(n-0)+\dfrac {1}{2}a(n^2-0^2)}$

$\displaystyle\longrightarrow\sf {\underline {\underline {s=v_0n+\dfrac {1}{2}an^2}}}$