Could Doc Brown's DeLorean Even Get Up To 88 MPH In That Parking Lot?

The DeLorean was known for a lot of things. Speed was not one of them. Underpowered even by early 1980s standards, it was a bizarre choice for a time machine that needed to hit 88 MPH to cross the temporal barrier. But could Doc Brown’s car even go that fast in the Twin, er, Lone Pine Mall parking lot like it does in the first Back to the Future? Opponaut Bullitt Ride did the math.

Doc Brown: If my calculations are correct... you'd be dead!

Back to the Future is hands down one of my favourite movie series, I’ve seen it dozens of times, and any time I see that it’s on TV I end up watching it... with the commercials... even though I could simply throw in the DVD to avoid the commercials... that must be the definition of lazy haha. As I’m sure it’s been discussed ad nauseam on the interweb, we are all aware that if Doc Brown’s calculations were correct Marty would have got to the line before the lightning struck. Fortunately, the Delorean was reliable enough to die on Marty just before the alarm clock went off...

So I was thinking this morning about Doc’s other “calculations”, mainly about getting up to 88mph in the parking lot of a mall. For the average sports car these days it’d probably be quite easy, however as we all know, the Delorean isn’t exactly a quick car, in fact it’s quite slow by today’s standards. So I wanted to try and do a quick rationalization as to whether a Delorean in real life would be able to reach 88mph in roughly the distances shown in the movie.

I did a quick Google search for the mall that the scene was shot in and I found this diagram that some other fans had already made for tourists showing the paths of the Delorean as well as the Libyans...

The paths on the diagram seem to be relatively accurate, so for the sake of this simple calculation I assumed that they were accurate. Okay, let’s look at Einstein’s drag race. In order to get the actual distance travelled I took a measurement off of Google maps and it was roughly 100 meters.


According to Wikipedia a Delorean will do 0-60mph in 8.8 seconds. To keep things simple I assumed a constant acceleration, up to 60mph this is a reasonable assumption, above 60pmh drag typically starts to take hold and acceleration starts to decrease, so if anything it’s an optimistic assumption. I’m also ignoring the fact the the parking lot in the movie was wet, and that the car would have likely been even slower with the added weight from the equipment Doc added to it. And yes, I’m also assuming that Doc hadn’t done anything to improve the acceleration performance of the Delorean.

The equation for acceleration is A=(V2-V1)/(T2-T1). In this case we know that V1=0, T1=0, V2=60mph, and T2=8.8seconds. Doing a unit conversion we end up with an acceleration of 1.894x10^-3 miles/second^2.


So knowing the acceleration and knowing that the car travelled roughly 100 meters or 6.214x10^-2 miles we can look at how long it would have taken to traverse that distance. For an object starting from a standstill the equation for distance assuming a constant acceleration is D=0.5AT^2. Solving that equation for time gives an answer of 8.1 seconds. What that means is that the Delorean wouldn’t even be going 60mph (let alone 88mph) by the time it reached Doc and Marty since it takes 8.8 seconds for the car to get to 60mph. Doc wasn’t joking about seeing some serious shit... him and Marty would be splattered across the Twin Pines parking lot!

Ok, well what about Marty? Did he have a better chance at making it since he had a rolling start? In the movie just before he rounds the corner and puts the hammer down we see a shot of the speedometer dropping down to below 35mph.


I’m going to assume that once he got the car straightened out from his little moment of oppositelock that he started his drag race at 35mph or 9.722x10^-3 miles/second. Another Google maps measurement shows that he had roughly 120 meters until whatever that little booth was that the Libyans smashed into.


The equation for final velocity assuming constant acceleration is V2^2 = V1^2 + 2A(X2-X1) where in this case X1=0 and X2=120 meters or 7.456x10^-2 miles. Plugging everything into the equation and solving for V2 gives an answer of approximately 70mph. Not that Doc had calculated this scenario... he was dead at this point, but Marty wouldn’t have had a shot at making it either. I’m sorry Jennifer, but if you want I’ll go up to the lake with you instead... and I’m talking about the original Jennifer here, not her lesser replacement.