Question

A force 3t2+2t+4:acts on a body of mass 2kg.the displacement of the body in first second is

Answer 1

Solution :

⏭ Given:

✏ Mass of body = 2kg

✏ Force = (3$\sf{t^2}$+2t+4) N

⏭ To Find :

✏ Displacement of the body in first second...

⏭ Concept:

$\bigstar \: \: \tt{ \red{F = m \times a}} \\ \\ \bigstar \sf \: \: \blue{v = \int a \: dt} \\ \\ \bigstar \sf \: \: \green{d = \int v \: dt}$

⏭ Calculation:

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$\implies \sf \: F = m \times a \\ \\ \implies \sf \: (3 {t}^{2} + 2t + 4) = 2 a \\ \\ \implies \: \boxed{ \tt{ \pink{a = \dfrac{3}{2} {t}^{2} + t + 2}}}$

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$\displaystyle\mapsto \sf \: v = \int \: a \: dt \\ \\ \mapsto \sf \: v = \int \:{ \huge{ ( }}\dfrac{3 {t}^{2} }{2} + t + 2{ \huge{)}} \: dt \\ \\ \mapsto \: \boxed{ \tt{ \green{ v = \frac{ {t}^{3} }{2 } + \frac{ {t}^{2} }{2} + 2t}}}$

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$\displaystyle \leadsto \sf \: d = \int \: v \: dt \\ \\ \leadsto \sf \: d = \int \:{ \huge{ ( }}\frac{ {t}^{3} }{2} + \frac{ {t}^{2} }{2} + 2t{ \huge{)}} \: dt \\ \\ \leadsto \sf \: d = \frac{ {t}^{4} }{8} + \frac{ {t}^{3} }{6} + {t}^{2} \\ \\ \red{ \sf\dag \: putting \: t = 1s} \\ \\ \leadsto \sf \: d = \frac{1}{8} + \frac{1}{6} + 1 \\ \\ \leadsto \sf \: d = 0.125 + 0.167 + 1 \\ \\ \leadsto \: \boxed{ \tt{ \pink{d = 1.292 \: m}}}$

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