Question

The force F exerted on a particle of mass 3kg is given by F=(9î+15j)N. If the particle starts from rest from the origin find its position at time 3 seconds.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Position=(13.5\hat{i}+22.5\hat{j})m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Mass \: of \: particle = 3 \: kg \\ \\ \tt: \implies Force = (9 \hat{i} + 15 \hat{j} )N \\ \\ \tt: \implies Time = 3 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Position \: at \: (t = 3 \: sec) =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies F = ma \\ \\ \tt: \implies F_{x} = m a_{x} \\ \\ \tt: \implies a_{x} = \frac{ 9 }{3} \\ \\ \green{\tt: \implies a_{x} = 3 \: {m/s}^{2} } \\ \\ \tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} a {t}^{2} \\ \\ \tt: \implies s_{x} = 0 \times 3 + \frac{1}{2} \times 3 \times {3}^{2} \\ \\ \green{\tt: \implies s_{x} = 13.5 \: m} \\ \\ \bold{Similarly : } \\ \tt: \implies a_{y} = \frac{ F_{y} }{m} \\ \\ \tt: \implies a_{y} = \frac{15}{3} \\ \\ \green{ \tt: \implies a_{y} =5 \: {m/s}^{2}} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{y} =0 \times 3 + \frac{1}{2} \times 5 \times {3}^{2} \\ \\ \green{\tt: \implies s_{y} =22.5 \: m} \\ \\ \green{\tt \therefore Position \: of \: particle \: at \: 3 \: sec \: is \: (13.5 \hat{i} + 22.5 \hat{j})m}$