The car will be traveling at 14 m/s at the end of the hill when it descends 32 meters, as the potential energy it loses is converted into kinetic energy.
the question of how fast a 1,700 kg car will be going at the end of a 32 m descent if friction has no effect in slowing its motion is c) 14 m/s.
we use the principle of conservation of energy. The potential energy lost by the car when it decreases in height (32 m) transforms into kinetic energy as the car accelerates. The potential energy (PE) can be calculated using PE = mgh, where m is mass, g is acceleration due to gravity (9.8 m/s2), and h is height. This gives us the kinetic energy (KE) at the bottom of the hill, and since KE = (1/2)mv2 where v is velocity, we can solve for v. Plugging in the numbers, we get (1,700 kg)(9.8 m/s2)(32 m) = (1/2)(1,700 kg)v2, which simplifies to v = 14 m/s.
by applying the conservation of energy principle and negating friction, we deduce that the car's speed at the bottom of the hill is 14 m/s.