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$A$, $B$ and $C$ are three points on a line of greatest slope of a smooth plane inclined at an angle of ${\theta ^ \circ }$ to the horizontal. $A$ is higher than $B$ and $B$ is higher than $C$, and the distances $AB$ and $BC$ are $1.76{\text{ }}m$ and $2.16{\text{ }}m$ respectively. A particle slides down the plane with constant acceleration $\alpha m{\text{ }}{s^{ - 2}}$. The speed of the particle at $A$ is $um{\text{ }}{s^{ - 1}}$ (see diagram). The particle takes $0.8{\text{ }}s$ to travel from $A$ to $B$ and takes $1.4{\text{ }}s$ to travel from $A$ to $C$. Find

a) the values of $u$ and $\alpha $,

b) the value of $\theta $.

پاسخ تشریحی :
نمایش پاسخ

a) $1.76 = 0.8u + 0.32\alpha $

$[1.76 + 2.16 = \left( {0.8 + 0.6} \right)u + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}{\left( {0.8 + 0.6} \right)^2}\alpha $ or

$2.16 = \left( {u + 0.8\alpha } \right)0.6 + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}{0.6^2}\alpha ]$

$3.92 = 1.4u + 0.98\alpha $ or $2.16 = 0.6u + 0.66\alpha $

$u = 1.4$ and $\alpha  = 2$

b) $\left[ {2 = 10\sin \theta } \right]$

$\theta  = 11.5$

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