Question

A certain cable car in San Francisco can stop in 10s when traveling at maximum speed. On one occasion, the driver sees a dog a distance d in front of the car and slams on the brakes instantly. The car reaches the dog 8.0s later, and the dog jumps off the track just in time. If the car travels 4.0m beyond the position of the dog before coming to a stop, how far was the car from the dog? (Hint: you will need three equations.)

Answer 1

hint:
assuming as maximum speed of a cable car, a speed of 9.5 mph, then we can write the subsequent formulas:
\[\Large \left\{ \begin{gathered}
d + {d_1} = {v_0}t - \frac{1}{2}a{t^2} \hfill \\
v = {v_0} - at \hfill \\
\end{gathered} \right.\]
where:
\[\Large {v_0} = 9.5\;mph \cong 15.3\;Km/h,{\kern 1pt} \quad {d_1} = 4\;meters\]