Centripetal Force

Consider a car is traveling 30 m/s on a road with lots of pot holes, and sharply turns to narrowly miss one. If the vehicle’s path is a 150 m curve radius and the force produced by its tires is 1,000 N, what is the car’s weight?

Hint
Centripetal acceleration:
$$$a=\frac{v^2}{r}$$$
where $$v$$ is the velocity, and $$r$$ is the radius.
Hint 2
$$$Force=mass\times acceleration$$$
Any motion on a curved path has accelerated motion:
$$$a=\frac{v^2}{r}$$$
where $$a$$ is centripetal acceleration, $$v$$ is the velocity, and $$r$$ is the radius. To stay on the curved path, a force directed towards the curvature’s center is required, and is called centripetal force:
$$$F=ma_{centripetal}=\frac{mv^2}{r}$$$
where $$m$$ is the mass and $$a$$ is the acceleration. Therefore,
$$$1,000N=\frac{m(30m/s)^2}{150m}$$$
Solving for mass:
$$$m=\frac{1,000N(150m)}{900m^2/s^2}=\frac{(1,000kg\cdot m)(150m)}{(s^2)(900m^2/s^2)}=167\:kg$$$
167 kg