Question

A curve in a highway has radius of curvature 409 m and is banked at 3.00°. On a day when the road is icy, what is the safest speed to go around the curve?

Answer 1

The safest speed to go around the curve is 14.49 m/s.

Step-by-step explanation:

Given that,

Radius of the circular curve, r = 409 m

Angle of banking,
`\theta=3^(\circ)`

We need to find the safest speed to go around the curve. On the banking of curve the centripetal force is balanced by the frictional force. The safest speed is given by :

`v=√(rg\ \tan\theta) \\\\v=√(409* 9.8\ \tan(3)) \\\\v=14.49\ m/s`

So, the safest speed to go around the curve is 14.49 m/s. Hence, this is the required solution.