In the question number 38, the speed of the particle at this time is

1.	In the question number 38, the speed of the particle at this time is
Answer» <html><body><p>16 m `s^(-1)` <br/>26 m `s^(-1)` <br/>36 m `s^(-1)` <br/><a href="https://interviewquestions.tuteehub.com/tag/46-317673" style="font-weight:bold;" target="_blank" title="Click to know more about 46">46</a> m `s^(-1)` </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Using eqn. (i) from question number 38, <br/> Velocity, `<a href="https://interviewquestions.tuteehub.com/tag/vecv-3257988" style="font-weight:bold;" target="_blank" title="Click to know more about VECV">VECV</a> = (dvecr)/(dt) = (d)/(dt) (<a href="https://interviewquestions.tuteehub.com/tag/5t-326357" style="font-weight:bold;" target="_blank" title="Click to know more about 5T">5T</a> +1.5t^(2)) hati + 1t^(2)hatj = (5+3t)hati +2thatj` <br/> At` = 6s, vecv = 23 hati + 12 hatj` <br/> The speed of the particle is `\|vecv\| = <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>((23)^(2) + (12)^(2)) = 26` m `s^(-1)`</body></html>

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