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  • Startin from initial point origin at t= 0, with the inintial velocity <img alt="10\; \hat{\mathrm{J}} \mathrm{m} / \mathrm{sec}" src="https://entrancecorner.oncodecogs.com/gif.latex?10%5C%3B%2

Startin from initial point origin at t= 0, with the inintial velocity 10\; \hat{\mathrm{J}} \mathrm{m} / \mathrm{sec} , a particle will moves in the x-y plane with a constant rate of change velocity of \mathrm{(15 \hat{\imath}+10 \hat{\jmath}) \mathrm{m} / \mathrm{sec}^2}. Find the coordinate at time 't' and also find the value of time 't'. if its x-coordinate is 20 m

Option: 1

\mathrm{\frac{82}{9 m}, \frac{2 \sqrt{2}}{3} \mathrm{sec}}


Option: 2

\mathrm{80 \mathrm{~m}, 5 \: \mathrm{sec}}


Option: 3

\mathrm{60 \mathrm{~m}, 4 \mathrm{sec}}


Option: 4

None of these

 


Answers (1)

best_answer

\begin{aligned} & \vec{u}=10 \hat{\jmath} \\ & \vec{a}=15 \hat{\jmath}+10 \hat{\jmath} \\ & s_x=v_{x \times t}+\frac{1}{2} 9 \times t^2 \\ & 20=0 \times t+\frac{1}{2} \times 15 \times t^2 \\ & t^2=\frac{40}{15}=\frac{8}{3} \\ & t=\frac{2 \sqrt{2}}{3} \mathrm{sec} . \end{aligned}\

\begin{aligned} s_y= & v_y \times t+\frac{1}{2} a y t^2 \\ y_0 & =5 \times \frac{2 \sqrt{2}}{3}+2 \times 10 \times\left(\frac{2 \sqrt{2}}{3}\right)^2 \\ & =\frac{10 \sqrt{2}}{3}+5 \times \frac{4 \times 2}{9} \\ & =\frac{10 \sqrt{2}}{3}+\frac{40}{9} \end{aligned}

\begin{aligned} y_0 & =\frac{30 \sqrt{2}+40}{9} \\ & =\frac{30 \times 1.4+40}{9} \\ & =\frac{82}{9} \mathrm{~m} . \end{aligned}

 

Posted by

Pankaj Sanodiya

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