Question

a motorcyclist moving with a velocity of 72 km per hour on a flat road takes a turn on the road at a point where the radius of curvature of road is 20 metres. The acceleration due to gravity is 10 m/s^2. In order to avoid skidding, he must not bend wrt the vertical plane by angle greater than

Answer 1

in order to avoid skidding ,
we should use the condition, $v\leq\sqrt{rgtan\theta}$
where v is the speed of motorcycle , r is the radius of curvature of road and $\theta$ is the angle made by motorcyclist with vertical plane.

Here Given,
v = 72 km/h = 72 × 5/18 = 20m/s
r = 20m and g = 10m/s²
now, $20\leq\sqrt{20\times10tan\theta}$
squaring both sides,
$400\leq200tan\theta$
$2\leq tan\theta$
$\theta\geq tan^{-1}(2)$

hence, he must not bent art the vertical plane by angle greater than tan^-1(2)

Answer 2

Answer:

Explanation:

we should use the condition,

where v is the speed of motorcycle , r is the radius of curvature of road and is the angle made by motorcyclist with vertical plane.

Here Given,

v = 72 km/h = 72 × 5/18 = 20m/s

r = 20m and g = 10m/s²

now,

squaring both sides,

hence, he must not bent art the vertical plane by angle greater than tan^-1(2)

Hope it helps !