Question

The radius of curvature of a road is 60 m .If the angle of banking is 27 degree , find the maximum speed with whose a car can be safely driven along the curv . ​

Answer 1

Answer:-

$\red{\bigstar}$ Maximum speed for safe banking

$\large\leadsto\boxed{\sf\pink{17.484 \: m/s}}$

• Given:-

Radius of curvature [r] = 60 m

Angle of banking = 27°

• To Find:-

Maximum speed for safe banking = ?

• Solution:-

We know,

$\blue{\bigstar}$$\sf\boxed{\red{tan \theta = \dfrac{v^{2}}{rg}}}$

here,

$\bf{\theta}$ = 27°

r = 60 m

g = 10 m/s²

Hence,

→ $\sf{tan 27 = \dfrac{v^{2}}{60 \times 10}}$

→ $\sf{tan 27 = \dfrac{v^{2}}{600}}$

→ $\sf{v^{2} = tan 27 \times 600}$ [tan 27° = 0.5095]

→ $\sf{v^{2} = 0.5095 \times 600}$

→ $\sf{v^{2} = 305.7}$

→ $\sf{v = \sqrt{305.7}}$

→ v = 17.484 m/s

Therefore, the maximum speed required for safe banking will be 17.484 m/s.