The stopping distance of a Tesla

Classical Mechanics Level 1

Have you ever wondered how far you could tailgate someone on the highway based on work-energy relation and a moving objects ability to stop at a certain distance? Well if not, then this problem will be right up your alley. Let us assume for a second that you just purchased the new Telsa Model S and you decide to go crazy on the highway and drive zero to 100 mph in 4s when suddenly a deer jumps out of the median. If this happens, how much distance do you need to come to a complete stop, so you don't hit the deer? Let us assume that the Tesla Model S has a mass of around 4,500 lbs. (recall: 1-pound equals 0.45 kg), moreover, the road has a road-to-tire friction coefficient of 0.8, the gravitational acceleration is $9.81\cdot m/{{s}^{2}}$ , and the brakes are operating correctly.

If we assume all these things, then what is the total stopping distance d (in units of feet)?

(hint: use the work-energy equation ${{W}_{f}}={{F}_{f}}\cdot d=\Delta KE$ , where $\Delta KE=\frac{1}{2}m(v_{f}^{2}-v_{i}^{2})$ , make sure you have the correct units, and solve for d)

401 ft 127 ft 407 ft 417 ft 365 ft

1 solution

Bradley Nordell
Jun 16, 2018

$\begin{aligned} & {{W}_{f}}={{F}_{f}}\cdot d=\Delta KE=-{{\mu }_{s}}mgd=-\frac{1}{2}mv_{0}^{2} \\ & d=\frac{v_{0}^{2}}{2{{\mu }_{s}}g}=\frac{{{(44.7m/s)}^{2}}}{2\cdot 0.8\cdot 9.81m/{{s}^{2}}}=127m=417ft \\ \end{aligned}$

The stopping distance of a Tesla

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