As a long-time aficionado of the original *Star Trek* series, it’s always exciting for me when I hear that Captain Kirk and Mr. Spock are going to make a reappearance on the big screen. Although it’ll be a bit strange without William Shatner and Leonard Nimoy running the show, what recourse is there? We’ve got the next generation playing the previous one.

Anyway, in the trailer we get a glimpse of the juvenile origins of the future Captain Kirk’s daredevil thrill-seeking persona, not to mention his incredible physical prowess. In the scene in question we see young James T. leap out of his classic convertible sports coupe moments before it projects itself off of a several-thousand-foot precipice. James saves himself by gripping the sandy ground and pulling himself to a stop just as he reaches the edge of the cliff.

Just for fun, we are going to apply a little physics and do a couple of quick back-of-the-envelope calculations to demonstrate how truly formidable the young Kirk really is. Our program will be to estimate a) the speed at which the car goes off the edge b) the speed of Kirk as he starts his slide towards the cliff and c) how much force he is going to have to apply to the road to bring himself to a stop at the brink. Now this one is going to be a little bit quantitative, so for all of you physics aficionados out there… here we go.

We see the car’s speedometer approaching 80 miles per hour (36 meters per second) in the trailer, so let’s estimate that the car is moving at that speed when it goes into its slide. The car appears to be about 30 meters from the edge when it starts skidding through soft dirt and sand. According to Newton’s Second Law we get:

`Fnet = Ffriction = µmg = ma`

where the acceleration of the car is completely due to the friction force. `M`

is the mass of the car, `g`

is equal to the acceleration due to gravity (9.8m/s2) , `µ`

is the coefficient of sliding friction between sand and tires (0.5 at most), and `a`

is the acceleration of the car. Solving for `a`

we get:

`a = µg = (0.5)(9.8m/s2) = 4.9 m/s2`

If we assume a relatively constant acceleration then

`a = (v2 - v02)/2x`

where `v0 = 36m/s, a = - 4.9 m/s2`

and `x = 30 m`

.

Solving for `v`

we find that the car goes over the cliff at a speed `v = 32 m/s`

.

Now the scene does a nice job of showing relative motion between the car, Jim, and the ground. When Kirk jumps off the car, although he is moving backwards relative to the car, he’s also still moving forward relative to the ground. A professional basketball player can jump with an initial liftoff velocity of few meters per second, so let’s generously give Kirk an initial takeoff speed of 4 m/s. But even so he’s still moving at a speed of 32m/s – 4m/s = 28 m/s towards the edge! And he’s only at most about 5 meters away when he jumps. In order to come to a stop in that distance Kirk will have to accelerate at a rate of

`a = (v2 - v02)/2x = (0 - 28m/s)2/2(5.0m) = 78 m/s2`

requiring a force of

`Fnet = Ffingers = ma`

and assuming that the young Kirk has a mass of around 50 kg we get

`Ffingers = ma = (50kg)(78m/s2) = 3900 Newtons (or 877 pounds)`

So what does all of this mean? It means that since the only thing(s) slowing him down are his fingers digging into the ground, James T. Kirk will have to exert a force of almost 900 pounds with those fingers to stop from being flung over the precipice. Although Dr. McCoy could never pull that one off, it is Captain Kirk after all. Or at least he will be Captain!

*Adam Weiner is the author of* Don’t Try This at Home! The Physics of Hollywood Movies.