Question

Turning a corner at a typical large intersection in a city means driving your car through a circular arc with a radius of about 25 m. if the maximum advisable acceleration of your vehicle through a turn on wet pavement is 0.40 times the free-fall acceleration, what is the maximum speed at which you should drive through this turn?

Answer 1

Answer:

9.89 m/s.

Explanation:

Given that,

The radius of the circular arc, r = 25 m

The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²

Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :

$a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{3.92\times 25} \\\\=9.89 m/s$

So, the maximum speed of the car should be 9.89 m/s.