a 915 kg car goes over a hill of circular arc. if the radius of the curve is 43 m, how fast can the car travel without leaving the road at the top of the arc?

Respuesta :

The car can travel at 20.5 m/s (74 km/hr). Any faster, and it will leave the road.

What is speed?

Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object travels along a path. In other words, velocity is a vector, whereas speed is a scalar value.

According to the question:

Two forces—gravity and the normal force of the road—are acting on the car as it crests the hill. These combined exert a net centripetal force, which will cause the car to travel in a circular arc.

[tex]$\vec{F}_c=\vec{F}_g+\vec{F}_N$[/tex]

Next, realize that "without leaving the road" actually refers to "just at the point where the car is going to leave the road" and that this corresponds to the condition FN = 0.

This means that our equation becomes

[tex]$F_c=F_g$[/tex]

and that gravity alone will determine the radius of the path the car will follow.

Substituting [tex]$F_c=\frac{m v^2}{r}$[/tex] and [tex]$F_g=m g$[/tex], we get

[tex]$\frac{m v^2}{r}=m g$[/tex]

Since we don't observe small, light automobiles flying airborne more frequently than larger ones, we can cancel the mass, which means that all cars traveling at this speed will follow the same course regardless of mass.

[tex]$\begin{aligned}& \frac{v^2}{r}=g \\& v=\sqrt{g r}\end{aligned}$[/tex]

After Substituting the values:

[tex]$v=\sqrt{(9.8)(43)}=20.5 m/s}$[/tex]

So, the car can travel at 20.5 m/s (74 km/hr).

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