The angle through which a cyclist bends when he covers a circular path of 34.3 m circumference in √22 sec is (g=9.8 m / s 2)

Question

The angle through which a cyclist bends when he covers a circular path of 34.3 m circumference in √22 sec is (g=9.8 m / s 2) (A)  15°  (B)  30°  (C)  60°  (D)  45°

Tardigrade Admin · Accepted Answer

Speed of particle v=(s/t)=(343/√22) m / s r=(s/2 π)=(34.3/2 π) m Now, tan θ=(v2/r g)=(((34.3)2/22)/(343)2 π × 9.8)=1 =45°

Q. The angle through which a cyclist bends when he covers a circular path of $34.3 m$ circumference in $22 sec$ is $(g = 9.8 m / s^{2})$

$1 5^{\circ}$

$3 0^{\circ}$

$6 0^{\circ}$

$4 5^{\circ}$

Speed of particle
$v = \frac{s}{t} = \frac{343}{22} m / s$
$r = \frac{s}{2 π} = \frac{34.3}{2 π} m$
Now, $tan θ = \frac{v ^{2}}{r g} = \frac{\frac{( 34.3 ) ^{2}}{22}}{\frac{343}{2 π} \times 9.8} = 1$
$= 4 5^{\circ}$

Q. The angle through which a cyclist bends when he covers a circular path of 34.3m circumference in 22​sec is (g=9.8m/s2)

15∘

30∘

60∘

45∘