find the equation of line passing through the given point and making the given angle with the given line( 2, -1), angle 45°, line 6 X + 5 Y - 1 =0
Answers
Answer:
EXPLANATION.
\begin{gathered} \sf : \implies \: a \: circular \: race \: track \: of \: 300m \: \\ \\ \sf : \implies \: banked \: at \: an \: angle \: of \: 15 \degree \\ \\ \sf : \implies \: coefficient \: of \: friction \: between \: the \: wheels \: of \: race \: car \: and \: road = 0.2 \\ \\ \sf : \implies \: \mu \: = 0.2\end{gathered}
:⟹acircularracetrackof300m
:⟹bankedatanangleof15°
:⟹coefficientoffrictionbetweenthewheelsofracecarandroad=0.2
:⟹μ=0.2
\begin{gathered} \sf : \implies \: \orange{{ \underline{1) = optimum \: speed \: of \: race \: car \: to \: avoid \: wear \: and \: tear \: on \: its \: tires \: }}} \\ \\ \sf : \implies \: \tan( \theta) = \frac{ {v}^{2} }{rg} \\ \\ \sf : \implies \: {v}^{2} = rg \tan( \theta) \\ \\ \sf : \implies \: {v}^{2} = 300 \times 10 \times \tan(15 \degree) \\ \\ \sf : \implies \: {v}^{2} = 300 \times 10 \times 0.26 \\ \\ \sf : \implies \: {v}^{2} = 780 \\ \\ \sf : \implies \: v \: = \sqrt{780} \approx \: 28\end{gathered}
:⟹
1)=optimumspeedofracecartoavoidwearandtearonitstires
:⟹tan(θ)=
rg
v
2
:⟹v
2
=rgtan(θ)
:⟹v
2
=300×10×tan(15°)
:⟹v
2
=300×10×0.26
:⟹v
2
=780
:⟹v=
780
≈28
\begin{gathered} \sf : \implies \: \orange{{ \underline{2) = maximum \: permissible \: speed \: to \: avoid \: slipping}}} \\ \\ \sf : \implies \: v_{m} \: = \sqrt{Rg \: ( \frac{ \mu \: + \tan( \theta) }{1 - \mu \: \tan( \theta) } }) \\ \\ \sf : \implies \: v_{m} \: = \sqrt{300 \times 10( \frac{0.2 \times \tan(15 \degree) }{1 - 0.2 \tan( 15 \degree) } } ) \\ \\ \sf : \implies \: v_{m} \: = \sqrt{3000( \frac{0.2 \times 0.26}{1 - 0.2 \times 0.26} } ) \\ \\ \sf : \implies \: v_{m} \: = \sqrt{1458.7} = 38.19 \: ms {}^{ - 1} \end{gathered}
:⟹
2)=maximumpermissiblespeedtoavoidslipping
:⟹v
m
=
Rg(
1−μtan(θ)
μ+tan(θ)
)
:⟹v
m
=
300×10(
1−0.2tan(15°)
0.2×tan(15°)
)
:⟹v
m
=
3000(
1−0.2×0.26
0.2×0.26
)
:⟹v
m
=
1458.7
=38.19ms
−1