- Home
- Standard 11
- Physics
2.Motion in Straight Line
hard
$x-$ अक्ष की दिशा में गतिमान एक कण के समय $t$ पर त्वरण $f$ को $f = f _{0}\left(1-\frac{ t }{ T }\right)$, समीकरण द्वारा व्यक्त किया जा सकता है, जबकि $f _{0}$ और $T$ नियतांक हैं। $t =0$ पर इस कण का वेग शून्य है। समय $t =0$ और उस क्षण के बीच अन्तराल में जबकि $f =0$ होगा, कण का वेग $(v_{ x })$ होगा
A$\frac{1}{2}{f_0}{t^2}$
B$\;{f_0}{T^2}$
C$\;\frac{1}{2}{f_0}T$
D$\;{f_0}T$
(AIPMT-2007)
Solution
$\begin{array}{l}
Give\,:\,At\,time\,t = 0,\,velocity,\,v = 0\\
Accelerations\,f\, = \,{f_0}\left( {1 – \frac{t}{T}} \right)\\
At\,f = 0,\,\,\,\,0 = {f_0}\left( {1 – \frac{t}{T}} \right)\\
\sin ce\,{f_0}\,is\,a\,constant,\\
\therefore \,1 – \frac{t}{T} = 0\,\,\,or\,\,t = T.
\end{array}$
$\begin{array}{l}
Also,\,acceleation\,f = \frac{{dv}}{{dt}}\\
\therefore \,\int\limits_0^{{v_x}} {dv} = \int\limits_{t = 0}^{t = T} {fdt = \int\limits_0^T {{f_0}} } \left( {1 – \frac{t}{T}} \right)dt\\
\therefore \,\,{v_x} = \left[ {{f_0}t – \frac{{{f_0}{t^2}}}{{2T}}} \right]_0^T = {f_0}T – \frac{{{f_0}{T^2}}}{{2T}} = \frac{1}{2}{f_0}T
\end{array}$
Give\,:\,At\,time\,t = 0,\,velocity,\,v = 0\\
Accelerations\,f\, = \,{f_0}\left( {1 – \frac{t}{T}} \right)\\
At\,f = 0,\,\,\,\,0 = {f_0}\left( {1 – \frac{t}{T}} \right)\\
\sin ce\,{f_0}\,is\,a\,constant,\\
\therefore \,1 – \frac{t}{T} = 0\,\,\,or\,\,t = T.
\end{array}$
$\begin{array}{l}
Also,\,acceleation\,f = \frac{{dv}}{{dt}}\\
\therefore \,\int\limits_0^{{v_x}} {dv} = \int\limits_{t = 0}^{t = T} {fdt = \int\limits_0^T {{f_0}} } \left( {1 – \frac{t}{T}} \right)dt\\
\therefore \,\,{v_x} = \left[ {{f_0}t – \frac{{{f_0}{t^2}}}{{2T}}} \right]_0^T = {f_0}T – \frac{{{f_0}{T^2}}}{{2T}} = \frac{1}{2}{f_0}T
\end{array}$
Standard 11
Physics
