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3-2.Motion in Plane
hard
A body is moving on a circle of radius $80 \,m$ with a speed $20 \,m / s$ which is decreasing at the rate $5 \,m / s ^2$ at an instant. The angle made by its acceleration with its velocity is ..........
A$45$
B$90$
C$135$
D$0$
Solution
(c)
Centrapetal acceleration, $a _{ c }=\frac{ v ^2}{ r }=\frac{20^2}{80}$
$=5\,m / s ^2$
If the brake is applied in the movement of body at $P$ of circular turn , acceleration along the target $=5 \,m / sec$
Angle between both the acceleration $=90^{\circ}$
the magnitude of resultant acceleration
$a =\sqrt{ a _{ c }^2+ a _t^2}=\sqrt{5^2+5^2}=5 \sqrt{2}\, m / sec$
List the resultant acceleration make an angle $\theta$ with tangent
$\tan \theta=\frac{ a _{ c }}{ a _5}=\frac{5}{5}=1$
$\theta=45^{\circ}$
Therefore required angle is $90^{\circ}+45^{\circ}$
$=135^{\circ}$
Centrapetal acceleration, $a _{ c }=\frac{ v ^2}{ r }=\frac{20^2}{80}$
$=5\,m / s ^2$
If the brake is applied in the movement of body at $P$ of circular turn , acceleration along the target $=5 \,m / sec$
Angle between both the acceleration $=90^{\circ}$
the magnitude of resultant acceleration
$a =\sqrt{ a _{ c }^2+ a _t^2}=\sqrt{5^2+5^2}=5 \sqrt{2}\, m / sec$
List the resultant acceleration make an angle $\theta$ with tangent
$\tan \theta=\frac{ a _{ c }}{ a _5}=\frac{5}{5}=1$
$\theta=45^{\circ}$
Therefore required angle is $90^{\circ}+45^{\circ}$
$=135^{\circ}$
Standard 11
Physics
