Question

the acceleration of a particle moving in a straight line is linearly with time t as bt. The particle started from origin with an initial velocity Vo .The distance travelled by particle in time t will be
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Answer 1

Given,

$\quad$

$\displaystyle\longrightarrow\sf {a=bt}$

$\quad$

where $\displaystyle\sf {b}$ is a constant.

$\quad$

But we know that $\displaystyle\sf {a=\dfrac {dv}{dt}.}$ Then,

$\quad$

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

$\quad$

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

$\quad$

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

$\quad$

$\displaystyle\longrightarrow\sf {\big[v\big]_{v_0}^v=b\left [\dfrac {t^2}{2}\right]_0^t}$

$\quad$

$\displaystyle\longrightarrow\sf {v-v_0=\dfrac {1}{2}bt^2}$

$\quad$

$\displaystyle\longrightarrow\sf {v=v_0+\dfrac {1}{2}bt^2}$

$\quad$

But, $\displaystyle\sf {v=\dfrac {ds}{dt}.}$ Then,

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {ds}{dt}=v_0+\dfrac {1}{2}bt^2}$

$\quad$

$\displaystyle\longrightarrow\sf {ds=\left (v_0+\dfrac {1}{2}bt^2\right)\ dt}$

$\quad$

$\displaystyle\longrightarrow\sf {\int\limits_0^sds=\int\limits_0^t\left (v_0+\dfrac {1}{2}bt^2\right)\ dt}$

$\quad$

$\displaystyle\longrightarrow\sf {\big [s\big]_0^s=v_0\big [t\big]_0^t+\dfrac {1}{2}b\left [\dfrac {t^3}{3}\right]_0^t}$

$\quad$

$\displaystyle\longrightarrow\sf {s-0=v_0(t-0)+\dfrac {1}{6}b(t^3-0^3)}$

$\quad$

$\displaystyle\longrightarrow\sf {\underline {\underline {s=v_0t+\dfrac {1}{6}bt^3}}}$