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An aeroplane is moving with velocity\mathrm{ \left(\sqrt{t}+\frac{2}{\sqrt{t}}\right)}, where t is time. When aeroplane at the maximum height the aeroplane will become stable. After some time rectum back to the runway with same velocity. what will be the acceleration at that particular time will be -

Option: 1

\mathrm{\frac{1}{\sqrt{t}}-\frac{1}{t^{3 / 2}}}


Option: 2

\mathrm{\frac{4}{\sqrt{t}}+\frac{1}{2 \sqrt{t}}}


Option: 3

\mathrm{\frac{1}{\sqrt{t}}+\frac{1}{t^{3 / 2}}}


Option: 4

\mathrm{\frac{1}{2 \sqrt{t}}-\frac{1}{t^{3 / 2}}}


Answers (1)

best_answer

Velocity of the aeroplane \mathrm{=\sqrt{t}+\frac{2}{\sqrt{t}}}

Acceleration of the aeroplane \mathrm{a=\frac{d}{d t}\left[\sqrt{t}+\frac{2}{\sqrt{t}}\right]}

                               \mathrm{ \begin{aligned} a & =\frac{d}{d t}\left[t^{1 / 2}+2 t^{-1 / 2}\right] \\ a & =\frac{1}{2} t^{-1 / 2}+2\left(-\frac{1}{2}\right) t^{-3 / 2} \\ \frac{d v}{d t}=a \Rightarrow & \frac{1}{2 \sqrt{t}}-\frac{1}{t^{3 / 2}} \end{aligned} }
                                                                                  Ans

Posted by

Gautam harsolia

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