A point moves along an arc of a circle of radius R- Its velocity depends upon the distance covered -s- as v -A-s- where a is constant- The angle - between the vector of total acceleration and tangential acceleration is-tan- -s-R-tan- -s-2R-tan- -2s-Rtan- -s-2R

The correct option is C tanθ =2s/R v=2√(s) .....(i) Squaring both sides v^2=a^2s ....(ii) a_c=v^2/R=a^2s/R a_t =dv/dt [multiply and divide RHS by ds] a ...

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Standard XII

Physics

Banking Angle

A point moves...

Question

A point moves along an arc of a circle of radius R. Its velocity depends upon the distance covered `s' as $v = A \sqrt{s}$ , where a is constant. The angle $θ$ between the vector of total acceleration and tangential acceleration is:

t a n θ = \sqrt{\frac{s}{R}}

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t a n θ = \sqrt{\frac{s}{2 R}}

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t a n θ = \frac{s}{2 R}

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t a n θ = \frac{2 s}{R}

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Solution

The correct option is D $t a n θ = \frac{2 s}{R}$

$v = 2 \sqrt{s}$ .....(i)
Squaring both sides
$v^{2} = a^{2} s$ ....(ii)
$a_{c} = \frac{v^{2}}{R} = \frac{a^{2} s}{R}$
$a_{t} = \frac{d v}{d t}$ [multiply and divide RHS by ds]
$a_{t} = \frac{d v}{d s} (\frac{d s}{d t}) = \frac{v d v}{d s}$ .....(iii) $[v = \frac{d s}{d t}]$
Differentiate equation(ii)
$2 v \frac{d v}{d s} = a^{2}$
$\frac{v d v}{d s} = \frac{a^{2}}{2}$ use this result in equation (iii) we get
$a_{t} = \frac{a^{2}}{2}$
$t a n θ = \frac{a_{c}}{a_{t}} = \frac{a^{2} s \times 2}{R a^{2}} = \frac{2 s}{R}$

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A point moves along an arc of a circle of radius R. Its velocity depends upon the distance covered `s' as v=A√s, where a is constant. The angle θ between the vector of total acceleration and tangential acceleration is:

The correct option is D tanθ=2sR v=2√s .....(i) Squaring both sides v2=a2s ....(ii) ac=v2R=a2sR at=dvdt [multiply and divide RHS by ds] at=dvds(dsdt)=vdvds .....(iii) [v=dsdt] Differentiate equation(ii) 2vdvds=a2 vdvds=a22 use this result in equation (iii) we get at=a22 tanθ=acat=a2s×2Ra2=2sR

A point moves along an arc of a circle of radius R. Its velocity depends upon the distance covered `s' as $v = A \sqrt{s}$ , where a is constant. The angle $θ$ between the vector of total acceleration and tangential acceleration is: