The stopping distance of cars

Assuming constant acceleration, we can use the kinematics equation $v^2 = v_0^2 + 2ad$ , where $v$ is final velocity, $v_0$ is initial velocity, $a$ is acceleration (deceleration in this case) and $d$ is displacement (which here is the braking distance.). In the given scenario, both cars have $v = 0$ , so our equation becomes

$\begin{aligned} 0 &= v_0^2 + 2ad \\ d &= \dfrac{\text{-}v_0^2}{2a} \end{aligned}$

Since $a$ is the same for both cars, we can say the braking distance is directly proportional to the square of the initial velocity; thus car B, which has an initial velocity three times as high as car A's, will require nine times as big a braking distance. So car B's braking distance will $9 \times 41 + 78 = \boxed{447 \text{m}}$

The kinetic energy of the car is proportional to its velocity squared. Thus, the answer is 41*(51/17)^2 + 78 = 447m.