মিটার পর্যন্ত) রৈখিকভাবে বৃদ্ধির ত্বরণ দেওয়া হয়

$$a = {5 \over {20}}x + 5 \Rightarrow a = {x \over 4} + 5$$

$$ \Rightarrow {{dv} \over {dt}} = {x \over 4} + 5$$ $$\because$$ $$\left( {a = {{dv} \over {dt}}} \right)$$

$$ \Rightarrow {{v\,.\,dt} \over {dx}} = {x \over 4} + 5$$ $$\because$$ $$\left( {{{dx} \over {dt}} = v \Rightarrow dt = {{dx} \over v}} \right)$$

$$ \Rightarrow \int\limits_5^v {v\,.\,dv = \int\limits_0^{20} {\left( {{x \over 4} + 5} \right)\,.\,dx} } $$

$$ \Rightarrow \left[ {{{{v^2}} \over 2}} \right]_5^v = \left[ {{{{x^2}} \over 8} + 5x} \right]_0^2$$

$$ \Rightarrow {{{v^2}} \over 5} - {{25} \over 2} = {{400} \over 8} + 100$$

$$ \Rightarrow {{{v^2}} \over 2} = 150 + {{25} \over 2} \Rightarrow {v^2} = 325$$

এখন, গতির দ্বিতীয়ার্ধে, ত্বরণ ধ্রুবক, যেমন a = 10 m/s²

$$\Rightarrow$$ v² $$-$$ u² = 2as

$$\Rightarrow$$ v'² $$-$$ v² = 2as (take, v = v', u = v)

$$\Rightarrow$$ v'² $$-$$ 325 = 2 $$\times$$ 10 $$\times$$ (35 $$-$$ 20) ($$\because$$ v² = 325)

$$\Rightarrow$$ v' = $$\sqrt {625} $$ = 25 m/s19:T4e7,During first half of the motion (i.e. upto x = 20 m) acceleration in increasing linearly is given by

$$a = {5 \over {20}}x + 5 \Rightarrow a = {x \over 4} + 5$$

$$ \Rightarrow {{dv} \over {dt}} = {x \over 4} + 5$$ $$\because$$ $$\left( {a = {{dv} \over {dt}}} \right)$$

$$ \Rightarrow {{v\,.\,dt} \over {dx}} = {x \over 4} + 5$$ $$\because$$ $$\left( {{{dx} \over {dt}} = v \Rightarrow dt = {{dx} \over v}} \right)$$

$$ \Rightarrow \int\limits_5^v {v\,.\,dv = \int\limits_0^{20} {\left( {{x \over 4} + 5} \right)\,.\,dx} } $$

$$ \Rightarrow \left[ {{{{v^2}} \over 2}} \right]_5^v = \left[ {{{{x^2}} \over 8} + 5x} \right]_0^2$$

$$ \Rightarrow {{{v^2}} \over

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