click on one answer only. compared to driving at 20 mph, about how much longer will it take for you to stop…

Explanation:

Step1: Recall kinetic - energy formula

The kinetic energy of a moving object is $K = \frac{1}{2}mv^{2}$, where $m$ is the mass of the object and $v$ is the velocity. The work - energy theorem states that the work done to stop the object is equal to its initial kinetic energy. Let the initial velocities be $v_1 = 20$ mph and $v_2=60$ mph.

Step2: Calculate the ratio of kinetic energies

The ratio of the kinetic energies $K_2$ and $K_1$ is $\frac{K_2}{K_1}=\frac{\frac{1}{2}mv_2^{2}}{\frac{1}{2}mv_1^{2}}=\frac{v_2^{2}}{v_1^{2}}$. Substitute $v_1 = 20$ mph and $v_2 = 60$ mph into the formula: $\frac{v_2^{2}}{v_1^{2}}=\frac{60^{2}}{20^{2}}=\frac{3600}{400}=9$. Since the work done to stop the vehicle (which is related to the stopping distance) is equal to the initial kinetic energy, the stopping distance at 60 mph is about 9 times the stopping distance at 20 mph.

Answer:

about nine times as far