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Velocity-v-time-t-and-displacement-x-of-a-particle-are-in-the-relation-1-t-3-v-3t-2-x-4t-If-at-t-0-x-1-m-then-find-displacement-at-t-1-s-

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Question Number 227034 by AgniMath last updated on 27/Dec/25
Velocity (v), time (t) and displacement (x) of a particle are in the relation (1 + t^3 )v + 3t^2 x = 4t. If at t = 0, x = 1 m then find displacement at t = 1 s.
$$\mathrm{Velocity}\:\left({v}\right),\:\mathrm{time}\:\left({t}\right)\:\mathrm{and}\:\mathrm{displacement} \\ $$$$\left({x}\right)\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{are}\:\mathrm{in}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\left(\mathrm{1}\:+\:{t}^{\mathrm{3}} \right){v}\:+\:\mathrm{3}{t}^{\mathrm{2}} {x}\:=\:\mathrm{4}{t}.\:\mathrm{If}\:\mathrm{at}\:{t}\:=\:\mathrm{0},\:{x}\:=\:\mathrm{1}\:\mathrm{m} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{displacement}\:\mathrm{at}\:{t}\:=\:\mathrm{1}\:\mathrm{s}. \\ $$
Commented by fantastic2 last updated on 27/Dec/25
convert to differrntial equation (1+t^3 )(dx/dt)+3t^2 x=4t ⇒(dx/dt)+((3t^2 )/(1+t^3 ))x=((4t)/(1+t^3 )) ∴x=((2t^2 +1)/(1+t^3 )) at t=1s x=((2×1+1)/(1+1))=(3/2)m
$${convert}\:{to}\:{differrntial}\:{equation} \\ $$$$\left(\mathrm{1}+{t}^{\mathrm{3}} \right)\frac{{dx}}{{dt}}+\mathrm{3}{t}^{\mathrm{2}} {x}=\mathrm{4}{t} \\ $$$$\Rightarrow\frac{{dx}}{{dt}}+\frac{\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{3}} }{x}=\frac{\mathrm{4}{t}}{\mathrm{1}+{t}^{\mathrm{3}} } \\ $$$$\therefore{x}=\frac{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{3}} } \\ $$$${at}\:{t}=\mathrm{1}{s} \\ $$$${x}=\frac{\mathrm{2}×\mathrm{1}+\mathrm{1}}{\mathrm{1}+\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{2}}{m} \\ $$

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