Question

A car of mass $1000kg$ negotiates a banked curve of radius $90m$ on a frictionless road. If the banking angle is $45°$, the speed of the car is:

Answer 1

Step 1: Given

Mass of the car, $m=1000Kg$

The radius of the banked road, $r=90m$

Banking angle, $\theta =45°$

Step 2: Determine the speed

The speed of the car on a banked road is given by,

Here, $\frac{m{v}^2}{r}$ is the centripetal acceleration acting on the car.

Under equilibrium,

$$\begin{gathered} \frac{m{v}^2}{r}\cos \theta =mg\sin \theta \\ v=\sqrt{rg}\tan \theta \\ =\sqrt{90\times 10}\tan \left(45\right) \\ =30m/s \end{gathered}$$

Therefore, the speed of the car is $30m/s$.